Đề thi toán học không biên giới năm 2016

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MATHEMATICS WITHOUT BORDERS 
2015-2016

WINTER 2016: GROUP 1

Problem 1. Which of the following symbols “<”, “>” or “=” should we place in the square, so that the 
equation would be true?

A) < B) > C) =

Problem 2. What is the missing number?

A) one B) two C) three

Problem 3. If

? =

A) +7 B) -7 C) +2

Problem 4. If

? =

A) 1 B) 2 C) 3

Problem 5. Replace the smileys with two of the cards in order to get the greatest possible sum.

What is the greatest sum?

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Problem 6. The number of leaves on a few three-leaf clovers can NOT be:

A) 11 B) 12 C) 15

Problem 7. How many of the following expressions are correct?

A) 1 B) 2 C) 3

Problem 8. Sonya has 2 fish. Amina has 2 fish more than Sonya. How many fish do Amina and Sonya 
have in total?

A) 4 B) 6 C) 8

Problem 9. There is a basket in a dark room. In the basket there are 2 yellow and 2 red apples. What is 
the smallest possible number of apples we would need to take out, without looking at their colour, in
order to ensure that we have taken out at least 1 red apple?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 B 1 + 1 + 3 > 6 – 2

2 В 4 + 2 = 6 Two is the missing number.
3 С 2 + 4 = 6; 6 – 6 = 0; 0 + 2 = 2

4 А + 7 = 10 ⇒ = 3
3 + = 7 ⇒ = 4
– = 4 – 3 = 1

5 С If we turn the second card upside down, we will get the greatest sum 9 + 7 = 16
6 А One three-leaf clover has 3 leaves; two three-leaf clovers have 6 leaves; three

clovers have 3+3+3=9 leaves; four clovers have 3+3+3+3=12 leaves; five
clovers have 3+3+3+3+3=15 leaves.

7 А Out of the three expressions only the first is correct.

8 В Amina has 4 fish, and together with Sonya they have 6 fish in total.
9 3 If the first two apples are yellow, the third would be red.

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SPRING 2016: GROUP 1

Problem 1.

A) 1 B) 2 C) 3

Problem 2. Which of the following is NOT true?

A) 9 + 1 + 2 = 12 B) 9 + 2 + 2 = 13 C) 9 + 2 + 4 = 14

Problem 3. Which of the following are one-digit numbers? 
0, 1, 2, 12, 3, 13 and 14

A) 7 B) 4 C) 2

Problem 4.

A) 1 B) 2 C) 3

Problem 5. By how much is 10 greater than 9?

A) 19 B) 1 C) 2

Problem 6. The sparrows on each tree are as many as the total number of trees. The total number of 
sparrows is 4. How many trees are there?

A) 2 B) 3 C) 4

Problem 7. I have 17 roses – white, yellow and red. The white and yellow roses together are 10, the
yellow and red roses together are 10. How many yellow roses are there?

A) 7 B) 3 C) 1

Problem 8.

A) 8 B) 9 C) 10

Problem 9. I have 1 apple, Yvette has 1 apple more than me, and Daria has 1 apple less than Yvette. 
How many apples in total do the three of us have?

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Problem 10. Nine children are playing hide and seek. One of them is seeking the others and finds 7 of

the children. How many of the children remain hidden?

A) 16 B) 2 C) 1

Problem 11. I added two different numbers and the sum I got is 2. Which of the two addends is greater?

Problem 12. Rather than deducting 1 from a certain number, I added 1 and got 2 as a result. What is the 
number I should have gotten?

Problem 13. Three identical balloons cost 10 cents more than a single balloon of the same kind. How 
many cents does a single balloon cost?

Problem 14. One addend is greater than 1 by 1 and the other addend is less than 2 by 2. What is the 
sum?

Problem 15. I wrote down all numbers from 5 to 16. How many are the digits used more than once?

Problem 16. John needs 3 more balloons in order to have 4. Peter has 1 balloon more than John. How 
many balloons do Peter and John have in total?

Problem 17.

= 1
+ Δ = 4
Δ + = 9

= ?

Problem 18. There are 6 flowers in a vase, each of which has 3 petals. I picked 4 of the flowers. What is 
the total number of petals on the remaining flowers?

Problem 19. A bunny eats 2 carrots every day. How many days will it take for the bunny to eat 6 
carrots?

Problem 20.

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 B ⇒

2 C A) true B) true C) NOT true 9+2+4=16

3 B The one-digit numbers are 0, 1, 2 and 3 ⇒ there are four in total.
4 B ⇒ ⇒

5 B

6 A If there is 1 tree, there would be 1 sparrow;

If there are 2 trees then the sparrows would be 2 + 2 = 4.
7 B ⏟

+ red = 17 ⇒ red = 7

yellow + ⏟ = 10 ⇒ yellow = 3 ⇒ There are 3 yellow roses.

8 B ⇒

9 A I have 1. Yvette has 1 + 1 = 2, and Daria has 2 – 1 = 1.
Together we have 1 + 2 + 1 = 4.

10 C Among the 9 children playing hide and seek, there are 8 hidden children. 7 of
them have been found.

The children that remain hidden are 8 – 7 = 1.

11 2 The addends are different according to the condition of the problem.
Therefore 2 can be presented as 2 + 0.

The greater of the two addends is 2.

12 0 I added 1 and got 2 as a result. Therefore the number that I added 1 to was 1.
The result I should have gotten is 1 – 1 = 0.

13 5 Balloon + balloon + balloon = balloon + 10 cents.
Therefore balloon + balloon = 10 cents.

One balloon costs 5 cents.

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15 3 I wrote down all numbers from 5 to 16: 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. 
The digits which have been used more than once are 1, 5 and 6. They are 3 in
total.

16 3 John needs 3 balloons in order to have 4. John has 1 balloon. Peter has 1 balloon
more than John. Peter has 2 balloons. Peter and John have

1 + 2 = 3 balloons in total.

17 6 =1

⏟ ⇒

⏟ ⇒ =6

18 6 There were 6 flowers, each of which had 3 petals. I picked 4 of the flowers. Now
there are 2 flowers in the vase with 3 petals each, i.e. the petals are 6.

19 3 A bunny eats 2 carrots every day. In two days it would eat 4 carrots. In three
days it would eat 6 carrots.

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FINAL 2016: GROUP 1

Problem 1.

A) 2 B) 1 C) 0

Problem 2. I have 17 roses – white, yellow and red. The white and yellow roses together are 10, the
yellow and red roses together are also 10. The roses of which colour are the least in number?

A) white B) yellow C) red

Problem 4. Adam, Bobby, Charles and Daniel won the top four places at a competition. Adam was 
ranked higher than Bobby, Charles was ranked lower than Daniel, and Bobby was ranked higher than
Daniel. Who came third?

A) Adam B) Bobby C) Daniel

Problem 4. Find the number in the following diagram:

A) 5 B) 6 C) 7

Problem 5. I will turn 15 in 8 years. How old was I 2 years ago?

A) 5 B) 6 C) 7

Problem 6. How many of the numbers 0, 1, 2, 3, 4 and 5 can be places in the empty square, so that the 
following equation + 5 < 9 will be true?

A) 3 B) 4 C) 5

Problem 7. What number will you get as a result of adding the numbers hidden under the shells?

1, 3, , 7, 9, , 13

A) 10 B) 14 C) 16

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A) 3 B) 4 C) 5

Problem 9. There are 3 teams participating in a football tournament. Each team played one game 
against each of the other teams. How many games have been played in total?

A) 3 B) 5 C) 6

Problem 10. There is a basket of apples in a dark room. Inside the basket there are 4 yellow and 2 red 
apples. What is the smallest number of apples we would need to take out (without looking), in order to

be sure that we have taken out at least two yellow apples?

A) 3 B) 4 C) 5

Problem 11. The sum of two one-digit numbers is 17. The smaller number was subtracted from the 
greater number. What is the difference?

Problem 12. The numbers 0, 1, 2, 7 and 10 are written down on a piece of paper. Annika erased two of 
the digits and the numbers which remained on the piece of paper were 0, 1 and 10. When she added
those numbers she got 11 as a sum. Pippi had her own piece of paper which also had the numbers 0, 1, 2,
7 and 10 written on it. She also erased two digits, correctly added the remaining numbers, but received a
sum smaller than that of Annika. What is the smallest possible sum that Pippi could have gotten?

Problem 13. It is well known that when a die is rolled, the winning number is the one found on top of 
the die (1, 2, 3, 4, 5 or 6).

When the die shown on the picture was rolled, the winning number was 3. Three dice were rolled and
there were three different winning numbers. The sum of the three numbers was 14. What is the smallest
winning number we got?

Problem 14. Find the next (Fifth) sum:

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Second sum:
0 + 1 + 1 = 2
Third sum:
0 + 1 + 1 + 2 = 4

Fourth sum:
0 + 1 + 1 + 2 + 4 = 8

Problem 15. John arranged 12 books on his shelf. The book Pippi Longstocking was arranged 8th from 
left to right. Which place would the same book be at if we were counting from right to left?

Problem 16. The number of one-digit numbers smaller than 5 is 1 less than the number of two-digit 
numbers smaller than . Find .

Problem 17. The numbers 1, 2, 3 and 4 must be placed in the following empty squares. Find the 
difference.

difference

Problem 18. Arnold and Mary have some pet fish. Mary has 2 fish more than Arnold. Together they 
have 18 fish. How many fish does Mary have?

Problem 19. All but 7 of a group of 18 children love ice cream. How many of the children do not love 
ice cream?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 А ⇒

2 B ⏟

+ red = 17 ⇒ red = 7

yellow + ⏟ = 10 ⇒ yellow = 3⇒ white = 7

There are 3 yellow roses.

3 С Adam was ranked higher than Bobby ⇒ the ranking is AB
Bobby was ranked higher than Daniel ⇒ the ranking is ABD
Charles was ranked lower than Daniel ⇒ the ranking is ABDC
Daniel came third.

4 B 4 + 9 = 13; 9 – 2 = 7.

Therefore 13 - 7= 6 ⇒ = 6

5 А I will be 15 in 8 years. I am currently 15 – 8 = 7 years old, and two years
ago I was 7 – 2 = 5 years old.

6 В 0 + 5 < 9, correct; 1 + 5 < 9, correct; 2 + 5 < 9, correct; 3 + 5 < 9, correct;
4 + 5 < 9, wrong; 5 + 5 < 9, wrong. + 5 < 9 is correct for 4 of the
numbers 0, 1, 2, 3, 4 and 5

7 C The numbers are 1, 3, 5, 7, 9, 11 and 13. The numbers hidden under the
shells are 5 and 11. Their sum is 16.

8 А Three different sums can be paid using three of the coins:
1 + 1 + 1 = 3; 1 + 1 + 5 = 7; 1 + 5 + 5 = 11

9 А If the teams are A, B and C, then the games played were A and B, B and
C, A and C. Three in total.

10 В In the worst case scenario, we would first take out the two red apples. In
this case the next two apples would definitely be yellow.

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12 3 On the piece of paper that has the numbers 0, 1, 2, 7 and 10 on it, Pippi
can erase the following:

10 - two digits 0 and 1; in which case the sum she would get is 10, which
is smaller than that of Anika.

7 and the first digit of 10; in which case the sum she would get is 3;
7 and the second digit of 10; in which case the sum she would get is 4.
2 and the first digit of 10; in which case the sum she would get is 8;
2 and the second digit of 10; in which case the sum she would get is 9.
1 and the first digit of 10; in which case the sum she would get is 9.
13 3 The possible options are: 14 = 6 + 6 + 2 = 6 + 5 + 3 = 6 + 4 + 4 = 5 + 5 +

4. The winning numbers are different only in the second sum. The
smallest winning number was 3.

14 16 The fifth sum is 0 + 1 + 1 + 2 + 4 + 8 = 16.

15 5 After this book, there would be 4 other books if we count from left to
right. If we count the books from right to left, the book would come after
those 4 books, therefore it would come 5th.

16 16 The one-digit numbers smaller than 5 are 5: 0, 1, 2, 3, 4. The two-digit
numbers smaller than 16 are 6: 10, 11, 12, 13, 14, 15.

The number that we would need to place in the square is 16.

17 2 ⏟

⏟

The difference is 2.

18 10 8 + 10 = 18, therefore Mary has 10 fish.

19 7 Out of 18 children, all but 7 love ice cream. Therefore 7 of them do not
love ice cream.

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TEAM COMPETITION – NESSEBAR, BULGARIA

MATHEMATICAL RELAY RACE

The answers to each problem are hidden behind the symbols @, #, &, § and * and are used in solving 
the following problem. Each team, consisting of three students of the same age group, must solve the
problems in 45 minutes and then fill a common answer sheet.

Problem 1. There are 14 chocolates in a box. Each of the three members of the team ate two chocolates. There are

now @ chocolates left in the box. Find @.

Problem 2. There are # sparrows perched on a bush. @ of them flew off the bush. The remaining 
sparrows are 4 less than those who flew off. Find #.

Problem 3. I have # yellow and red flowers. Seven of them are tullips, and the rest are roses. Two of the 
flowers are yellow and the rest are red. What is the smallest possible number of red roses? Mark your
answer by &. Find &.

Problem 4. Two identical chocolate bars cost as much as & identical sweets. Six chocolate bars cost as 
much as § sweets. Find §.

Problem 5. The two-digit numbers smaller than * are §. Find *.

ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 @ = 8 2+2+2=6, therefore 6 chocolates have been eaten
already. There are 14-6=8 chocolates left.

2 # = 12 The sparrows left on the bush are 8-4=4. At first
the sparrows were 8+4=12.

3 & = 3

The roses are 12-7=5. In order to get the smallest
possible number of red roses, both yellow flowers
must be roses. 5-2=3, therefore at least 3 of the
roses are red.

4 § = 9

2 chocolate bars + 2 chocolate bars + 2 chocolate
bars are equal to 3 sweets + 3 sweets + 3 sweets =
9 sweets.

5 * = 19

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MATHEMATICS WITHOUT BORDERS 
2015-2016

AUTUMN 2015: GROUP 2

Problem 1. What is the missing number?

A) 10 B) 11 C) 21

Problem 2. The sum of 10 + 8 equals:

A) the sum of 6 and 11 B) the difference of 14 and 4 C) the sum of 9 and 9

Problem 3. In a sum of two numbers, one addend is greater than 2 by 2, while the other addend is 
smaller than 1 by 2. The sum is:

A) 2 B) 4 C) 5

Problem 4. What is the largest two-digit number with 0 as a units digit?

A) 10 B) 90 C) 100

Problem 5. How many of the following expressions are correct? 
11-2 > 13

18+3 > 20
12-5 = 3+4

A) 1 B) 2 C) 3

Problem 6. How many are all the possible digits that can be placed instead of @, so that 
would be true?

A) 9 B) 6 C) 3

Problem 7. What is the largest sum of 2 different single-digit numbers?

A) 19 B) 18 C) 17

Problem 8. I thought of a number. I added it to 2 and got 10. The number I thought of is:

A) 12 B) 8 C) 10

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A) 11 B) 19 C) 21

Problem 10. How many are the two-digit numbers that do NOT have 9 as a ones digit?

A) 9 B) 81 C) 90

Problem 11. Peter solved 3 problems, Iva solved 2 problems less than Peter; Mary solved one problem 
more than Iva. How many problems did Mary solve?

Problem 12. There is a basket in a dark room. In the basket there are 2 yellow and 3 red apples. What is 
the smallest possible number of apples we would need to take out, without looking at their colour, in
order to ensure that we have taken out 2 red apples?

Problem 13. How many single-digits numbers is the magic square made of?

6 8 1

5
2

Problem 14. How many sheets of paper are there between the third and the seventh pages of a book?

Problem 15. Find the sum of all two-digit numbers whose sum of digits is 3?

Problem 16. How many numbers have been omitted in the sequence 1, 11, 21, 31, ..., 81, 91?

Problem 17. Joel has a few bunnies. Each one of them has 2 ears and 4 legs. If their ears are 10 in total, 
how many legs do they have in total?

Problem 18. If the minuend is 9 and the subtrahend is 9, we get a difference of?

Problem 19. How many units are there in the number equal to

– – – – – ?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 B 
2 C 10 + 8 = 18, 18 = 9 + 9

3 C 2 + 2 = 4; 2 – 1 = 1
4 + 1 = 5

4 B 90

5 B 9>13; 21>20; 7=7
6 C 36<37; 36<38; 36<39

7 C 9 + 8 = 17

8 B ? + 2 = 10 ? = 8

9 B 10; 10 – 1 = 9; 10 + 9 = 19

10 B ⏟ ⏟ …., ⏟

9+9+9+9+9+9+9+9+9=81

11 2 Iva solves 1 problem, Maria solved 1 + 1 = 2 problems.

12 4 If we were to take both yellow apples, the next 2 would be red. Therefore
if we take 4 apples, there will always be 2 red apples among them.

13 8 6 8 1

0 5 10

9 2 4

14 1 This is the list of paper with page numbers 5 and 6.
15 63 The numbers are 12, 21 and 30. Their sum is 63.
16 4 The numbers 41, 51, 61 and 71 have been skipped.

17 20 There are 10 ears. Therefore the bunnies are 5. Each bunny has 4 legs.
4 + 4 + 4 + 4 + 4 = 20.

18 0 9 – 9 = 0

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WINTER 2016: GROUP 2 
Problem 1. What is the missing number?

A) 1 B) 2 C) 3

Problem 2. The sum of is:

A) 90 B) 80 C) 70

Problem 3. In a sum of two numbers, one of the addends is greater than 20 by 20, and the other addend 
is smaller than 20 by 10. The sum of the two numbers is:

A) 50 B) 40 C) 30

Problem 4. How many of the following expressions are correct?

A) 1 B) 2 C) 3

Problem 5. What is the missing number „?”?

A) 8 B) 18 C) 35

Problem 6. How many digits can we place instead of @, so that would not be true?

A) 10 B) 9 C) 1

Problem 7. What is the greatest sum of 3 different one-digit numbers?

A) 23 B) 24 C) 25

Problem 8. There is a basket in a dark room. In the basket there are 6 yellow and 5 red apples. What is 
the smallest possible number of apples we would need to take out, without looking at their colour, in
order to ensure that we have taken out at least 3 red apples?

A) 8 B) 9 C) 10

Problem 9. If we add the number equal to 94 – (46 + 38) to the number equal to 94 – 46 +38, what
result would we get?

A) 86 B) 76 C) 96

Problem 10. A gallery has 96 paintings. 32 of them were sold on the first day, and on the second day the 
gallery sold 3 paintings more than the previous day. How many paintings are still not sold?

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Problem 11. Three friends weigh respectively 24, 30 and 42 kilograms. They want to cross a river by 
using a boat that can carry a maximum of 70 kg. At least how many times would this boat need to cross
the river, so that all three of them would get to the opposite shore

Problem 12. How many tens are there in the number equal to

– – – – – ?
Problem 13. What is the greatest number in the magic square?

6 8 1

Problem 14. In how many squares can you find the letter A?

Problem 15. Place the digits 1, 2, 3 and 4 in the squares in a way that would result in the greatest sum. 
. What is the sum?

Problem 16. Boko and Tsoko went fishing with their sons. All of them caught an equal number of fish. 
How much fish did each of them catch, if they caught 9 fish in total?

Problem 17. The minuend is greater than the subtrahend by 2. What is the difference?

Problem 18. How many are the three digit numbers different from 102, that can be derived from the 
number 102 by randomly moving the digits of the number around?

Problem 19. If we follow the rule:

then which number do we need to place in the square with the ant in it?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 A 57 – ? = 56; ? = 1

2 А 90

3 А One of the addends is 20 + 20 = 40, and the other is 20 – 10 = 10. The sum is 
50.

4 B 40 – 2 = 38, i.e. the first expression is not correct. The next two expressions are 
correct.

5 А The missing number in the circle is 35. Then we must add 8 to the number 35,
in order to get 43.

The number we are looking for is 8.

6 А We need to find out the following: for how many digits @ is it NOT true that:
40 > 4@?

For all ten digits: 0, 1, ..., 9.
7 В 9 + 8 + 7 = 24

8 B In the worst case scenario, we would take out all of the yellow apples first.
Then after 3 more attempts, we would have taken out 3 red apples, i.e. 9 in
total.

9 С The first addend is 10, and the second is 86. The sum is 96.

10 С The paintings sold on the second day were 35. The paintings sold on the first
and second day together are 67.

The paintings that remain unsold are 96 – 67 = 29.

11 3 Let C denotes the heaviest of the three friends, A - the lightest one, and B - the 
third one.

It would be impossible for all three of them to cross the river in one go,
because 24 + 30 + 42 = 96 > 70.

Therefore the boat would have to return at least once, and the smallest possible
number of river crossings would be 3.

Following is an example of a way in which all three friends can cross the river
to the opposite shore:

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A crosses back to the initial shore.

A and C now cross to the opposite shore together.

12 10 – – – – –
= 20 + 20 + 40 + 20 + 0 = 100. In the number 100 there are 10 tens.
13 10 The magical sum is 15.

The numbers in the second row are 0, 5 and 10, and in the third row are 9, 2
and 4.

The greatest number is 10.

14 4 The letter A is in one square 1 1, in two squares 2 2 and in one square 3 3.
15 46 1 + 2 + 43 = 46

16 3 or 1 If we assume that the problem speaks of four people – two fathers and two
sons, then the result would be impossible, because 9 is not divisible by 4.

Therefore the problem must speak of three people: a grandfather, his son, and
his grandson, or of 9 people: two fathers and seven sons.

17 2  + 2 –  = 2

18 3 The numbers are 102, 120, 201 and 210.
One of them has been written down already.
19 0 The numbers are as follows:

At the bottom: 9, 5, 2, 0
Above: 4, 3, 2

Above: 1, 1

And the number at the top is 0.

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SPRING 2016: GROUP 2

Problem 1 If –  then  =

A) 100 B) 99 C) 98

Problem 2. Which of the following lengths is the shortest?

A) 3 mm B) 2 cm C) 1 dm

Problem 3. If  , then  =

A) 0 B) 9 C) 8

Problem 4. I chose a random number. I switched the numbers of ones and tens. I added 19 to the 
resulting number and got 24. What is the number I had originally chosen?

A) 15 B) 50 C) 51

Problem 5. Alia and Daniel had 24 sweets at first. Then Alia bought 2 more sweets and she now has 12 
sweets more than Daniel. How many sweets does she have at the moment?

A) 18 B) 19 C) 20

Problem 6. The even numbers from 3 to , inclusive, is 20. What is the greatest possible value of ?

A) 41 B) 42 C) 43

Problem 7. Which of the following numbers is the smallest?

A) 3 + 2 2 B) 13 – 3 1 C) (3 + 2) 2

Problem 8. The number of sparrows on each tree is equal to the total number of trees. The total number 
of sparrows is 16. How many trees are there?

A) 3 B) 4 C) 5

Problem 9. Two two-digit numbers have been written using 4 different digits. Which of the following 
sums is possible?

A) 22 B) 26 C) 33

Problem 10. I bought 9 stamps, worth 6 cents each, and I payed using 6 coins of 10 cents. In how many 
different ways can I get my change?

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Problem 11. The numbers 1, 2, 3, 4 and 6 are written down on two pieces of paper. The product of the 
numbers from one of the pieces is equal to the product of the numbers from the other piece. How many
numbers are there on the piece of paper that has the number 1?

Problem 12. There are 2 grandmothers, 4 mothers, 4 daughters and 2 granddaughters in a room. What's 
the smallest possible number of people in that room?

Problem 13. There are 22 students in a class. Twelve of the students have the highest grade in less than 
four subjects, and 12 have the highest grade in more than two subjects. How many students have the
highest grade in exactly three subjects?

Problem 14. In Rose’s garden there are 88 roses which are not in bloom yet and 8 which are blooming. 
Every day 4 new roses bloom and the ones that are already blooming do not fade. How many days will it
take for the blossoming and non-blossoming roses to be an equal number?

Problem 15. Replace the smileys with two of the cards in order to get the greatest possible product.

What is the greatest possible product?

Problem 16. The square is ‘magical’. Calculate the number A.

21 18
27 15 А

Problem 17. If ⏟

⏟ 

, then  = ....

Problem 18. The product of five numbers is 5. What is their sum?

Problem 19. A container full of water weighs 20 kg and when half full it weighs as much as 3 empty 
containers. How many kilograms does this container weigh when it is empty?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 B 100 –(29 + 37) =  – 65, then 100 – 66 =  - 65 34 =  - 65  = 99

2 A A) 3 mm B) 2 cm = 20 mm C) 1 dm = 10 cm = 100 mm 
3 C If  , then 

17 = 25   = 8

4 B The number with exchanged digits of the ones and tens is
24 – 19 = 5. Therefore the originally chosen number is 50.
From 50 we can get 05 = 5 and 5 + 19 = 24.

5 B Before buying the extra sweets, Alia had 10 sweets more than Daniel.
24 – 10 = 14 and 14 2 = 7, therefore before buying the extra sweets Alia
had 17 sweets and Daniel had 7. At the moment Alia has 19 sweets.

6 C The 20 even numbers from 3 onwards are 4, 6, 8, 10, 12, 14, 16, 18, 20,
22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42.

The even numbers from 3 to 42, inclusive, are 20.
The even numbers from 3 to 43, inclusive, are 20.
The even numbers from 3 to 44, inclusive, are 21.

7 A 3 + 2 2 = 3 + 4 = 7; 13 – 3 1 = 13 – 3 = 10; (3 + 2) 2 = 10
8 B If the trees are 3, the sparrows would be 3 3 = 9;

If the trees are 4, the sparrows would be 4 4 = 16;
If the trees are 5, the sparrows would be 5 5 = 25.

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10 C The change would be equal to 6 10 – 9 6 = 6 cents. It can be given in
5 different ways:

5 + 1 = 2 + 2 + 2 = 2 + 2 + 1 + 1= 2 + 1 + 1 + 1 + 1 = 1 + 1 + 1+ 1 +1+1.

11 3 The product of the numbers is equal to 1 2 3 4 6 = 144. Therefore
we would need to write numbers that have a product of 12 on the pieces of
paper.

The numbers can be written down as follows: 1, 3 and 4 on the first piece
of paper, 2 and 6 on the second piece of paper, or 3 and 4 on the first piece
of paper, 1, 2 and 6 on the second piece of paper. The pieces of paper that
has the number 1 on it has 3 numbers written on it.

12 6 In order for one of the women to be a grandmother, she would need to
have a daughter, and a granddaughter. Therefore if there are two

grandmothers, who are also mothers, they have one daughter each, i.e. 2
daughters, each of whom is also a mother to 1 granddaughter – 2

granddaughters, who are also daughters.
The two granddaughters are also 2 daughters.

There are now 2 daughters left, who are also 2 mothers.
There are now 2 mothers left, who are also 2 grandmothers.

13 2 The total number of students in the class plus the number of students who
have the highest grade in 3 subjects equals 12 + 12 = 24. If we calculate
24 - 22 we would get the number of students who have the highest grade
in 3 subjects, i.e. 2.

14 10 The roses in blossom and those not yet in blossom are 96 in total. The
number of roses in blossom must increase by 96 2 – 8 = 40 roses. That
can happen in 40 4 = 10 days.

15 63 The possible products are 2 6; 2 7; 6 7; 2 9; 7 9. The greatest
among them is 63.

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16 3 We can find the answer by comparing the sums of the numbers from the
first column (B, 27, C) to the diagonal (B, 15, 24).

They are equal, B + 27 + C = B + 15 + 24, therefore 27 + C = 39.
We get that C = 12, therefore the ‘magical’ sum is 45 (12 + 15 + 18).
27 + 15 + A = 45, therefore A = 3.

17 5 If ⏟

⏟



then 20 = 4   = 5.
18 9 5 = 5 1 1 1 1,

therefore the sum we are looking for is 5 + 1 + 1 + 1 + 1 = 9.

19 4 The weight of the water in a half-full vessel is equal to two empty vessels.
The weight of the water in a full vessel weighs as much as 4 empty

vessels. The weight of the vessel plus the water inside it is equal to 5
empty vessels.

Therefore one empty vessel would be equal to 20 5 = 4 kg.

20 2 A B C D

A + + +

B + – +

C + – –

D + + –

If we add the number of hand shakes, the number must be divisible by 2,

because each hand shake is counted twice.

In this case the number of hand shakes is 6 + x.

We can mark the number of David’s handshakes with x. The number x 
can NOT be greater than 3.

6 + x can be divided by 2 only if x is either 0 or 2.

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FINAL 2016: GROUP 2

Problem 1. The product of all even one-digit numbers that are divisible by 3, is:

A) 0 B) 6 C) 18

Problem 2. What number should be placed instead of  so that the following equation would be true? 


A) 35 B) 36 C) 37

Problem 3. When Adam was counting the numbers from 1 to 50, he got distracted and he forgot to 
count the numbers that are divisible by 2 or by 3. How many numbers, smaller than 31, did he forget to
count?

A) 20 B) 10 C) 5

Problem 4. There are 20 odd numbers from 2 to , inclusive. What is the greatest possible value of ?

A) 41 B) 42 C) 43

Problem 5. Adam, Bobby, Charles and Daniel won the top four places at a competition. Adam was 
ranked higher than Bobby, Charles was ranked lower than Daniel, and Bobby was ranked higher than
Daniel. Who came third?

A) Adam B) Bobby C) Daniel

Problem 6. A container full of water weighs 21 kg and when half full it weighs as much as 4 empty 
containers. How many kg of water are there in the container when it is full?

A) 3 B) 16 C) 18

Problem 7. When I grow 8 years older than I am now, I will be twice as old as my brother who was 
born 2 years ago. How old am I at the moment?

A) 10 B) 12 C) 20

Problem 8. How many numbers can we place in the empty square, so that the following equation would 
be true?

 5 < 25?

A) 4 B) 5 C) more than 5

Problem 9. By how much is the number hidden under the first shell smaller than the number hidden 
under the second shell?

4, 7, 13, , 34, , 67

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Problem 10. We are given the numbers 1, 2, 3 and 4. If we erase two of them, then the product of the 
remaining numbers can be presented as the product of two equal multipliers. Which numbers should we

erase to do that?

A) 1 and 2 B) 2 and 3 C) 2 and 4

Problem 11. A few football teams are participating in a tournament. After a game has been played, only 
the winner moves forward into the tournament. If the teams are 16, what is the minimum number of
games that must be played in order for one of the teams to become a champion?

Problem 12. There is a basket of apples in a dark room. There are 4 yellow and 2 red apples inside it. 
What is the minimum number of apples you would need to take out (without looking) in order to be sure
that you have taken out 2 yellow and 1 red apples?

Problem 13. The sum of 11 one-digit numbers is 98. What is the smallest among these numbers?

Problem 14. On the figure below you can see that in the middle there is a square with a side of 1cm. On 
each of its sides there is another square, each with sides of 1cm. On each of the sides of the newly
formed figure, there is one extra square with a side of 1cm. How many squares are there in total on the
figure?

Problem 15. Here is what a few children said about the number 63: 
Adam: “This is a number made up of odd numbers!”

Bryan: “This number is a product of the numbers 7 and 9!”
Steve: “This number has 63 units!”

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Problem 16. I bought 9 sweets, each of which costs 7 cents, and I paid using 7 coins of 10 cents. In how 
many different ways can the shopkeeper give me my change?

Problem 17. There are 26 students in a class of second-graders. 15 of them have less than four balloons, 
and 17 have more than two balloons. How many of the students have more than three balloons?

Problem 18. What is the smallest possible sum of the numbers that we would need to place in the 6 
empty squares, so that the sum of the numbers in order of rows, diagonals, and columns would be the
same?

2
2
2

Problem 19. The digits used to write down the even two-digit numbers are more than the digits used to 
write down the odd one-digit numbers. By how many?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 А The even one-digit numbers divisible by 3 are 0 and 6.
Their product is 0.

2 А We can write down the equality as follows:

3 + 15 + 35 + 63 = 4 + 16 + 25 + 36 +   = 35

3 А

He forgot to count all even numbers, of which there are 15, as well as all odd
numbers divisible by 3, which are 3, 9, 15, 21 and 27.

He forgot to count 20 numbers in total.

4 В

The 20 odd numbers from 2 onwards are 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23,
25, 27, 29, 31, 33, 35, 37, 39,41.

The odd numbers from 2 to 41, inclusive, are 20.
The even numbers from 2 to 42, inclusive, are 20.

5 C The four boys are ranked as follows: AB and DC, therefore ABDC.

6 С

If the half-full container weighs as much as 4 empty containers, then the
weight of the water in a half-full container is equal to the weigh of 3 empty
containers. The weight of the water in a full container is equal to the weight
of 6 empty containers. I.e. when full of water, the container weighs as much
as 7 empty containers. Therefore one empty container weighs
The water in a full container weighs 21 – 3 = 18 kg.

7 В

We can find the correct answer by checking each possible answer. If I am 10
years old now, then my brother is 2 years old. In 8 years I will be 18 and my
brother will be 10. The number 18 is not twice as big as 10. If I am 12 now,
my brother is 2. In 8 years I will be 20 and he will be 10. This is the correct
answer.

8 В All numbers from 0 to 4 (5 numbers).

9 С

4, 7, 13, , 34, , 67

7 = 4 + 1 3;
13 = 7 + 2 3;

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The next number is 34 + 5 3 = 49,
The next number is 49 + 6 3 = 67.

The difference we are looking for is 49 – 22 = 27.

10 В If we were to erase the numbers 2 and 3, we would get a product of 4, which
we can present as 2

11 15

First we split the 16 teams into 8 couples.

They play 8 games, therefore there are 8 winners.
8 teams carry on to the second round.

8 teams play 4 games in the second round.

4 teams carry on to the third round, to play 2 games.
Final: 2 teams play 1 game.

The games played in total are 8 + 4 + 2 + 1 = 15.

12 5 In the worst case scenario we would take out all 4 yellow apples first, and the
5th apple would be red.

13 8 The sum of 11 one-digit numbers can be 99 at most. In this case it is 98.
Therefore, one of them is 8.

14 18 There are 14 squares with a side of 1 cm on the figure; 4 squares with a side 
of 2 cm and 1 square with a side of 3 cm. There are 18 squares in total. 
15 2 Only Adam’s claim is not true.

16 6

The change is 70-63=7 cents.

I can get my change in 6 different ways:
7 coins of 1 cent;

5 coins of 1 cent + 1 coin of 2 cents;
3 coins of 1 cent + 2 coins of 2 cents;
2 coins of 1 cent + 1 coin of 5 cents;
1 coin of 1 cent + 3 coins of 2 cents;
1 coin of 2 cents + 1 coin of 5 cents.

17 11 15+17-26=6 children have 3 balloons each. 17-6=11 students have more than
3 balloons each.

18 3

1 2 0
0 1 2 
2 0 1

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We can compare the sums of the numbers from the first row and from the
first column. They are equal, therefore they must have two more equal
numbers each - х.

2 x 
x 2 
2

We can then compare the second row and the second column:
2 x

x 2 
2 x

We can then compare the first row with the second column and the first row
with the third column and we would get the following:

y 2 x 
x y 2 
2 x y

If we then compare the sums along the diagonals, we will get that

y + у = 2 + x the smallest possible value is y=1 and x=0.

The sum we are looking for is 3.

19 85 or 5 There are 45 even two-digit numbers and they have been written down using
90 digits. The odd one-digit numbers are written down using 5 digits. The
answer we are looking for is 90 – 5 = 85.

Another answer is also possible:

The even two-digit numbers are written down using 10 digits and the odd
one-digit numbers are written down using 5 digits. In this case the answer
would be 10 – 5 = 5.

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TEAM COMPETITION – NESSEBAR, BULGARIA

MATHEMATICAL RELAY RACE

GROUP 2

Problem 1. The number of two-digit numbers that can be presented as a product of two consecutive 
numbers is @. Find @.

Problem 2. If the dividend is ⏟

, and the divisor is 7, what is the quotient #?

Problem 3. Little Red Riding Hood needs to cross a river by going through the only bridge, in order to 
get to her grandmother’s village. She can reach the bridge using & different roads, and she can use two 
different roads from the bridge to her grandmother’s village. It turns out she can reach her grandmother’s
village using # different routes. Find &.

Problem 4. Bugs Bunny loves eating cabbage and carrots. He eats either &+1 carrots or 4 cabbages 
every day. In one week Bugs Bunny ate 30 carrots and § cabbages. Find §.

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 @ = 7

2 3 = 6 < 10;

3 4 = 12; 4 5= 20; 5 6 = 30; 6 7 = 42;
7 8 = 56; 8 9 = 72; 9 10 = 90;

10 11= 110 >99.

The number we are looking for is @ =7. 
2 # = 8 The dividend is 2 + 4 + 6 + 8 + 10 + 12 + 14.

The quotient is 56 7=8.
3 & = 4 & 2 = # = 8 & = 4.

4 § =4

Bugs Bunny eats 5 carrots a day. It would take him 6 days to eat
30 carrots. He would only eat cabbage on the seventh day – he
would have to eat 3 cabbages.

§=4.

5 * = 3

We found that §=4.

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MATHEMATICS WITHOUT BORDERS 
2015-2016

AUTUMN 2015: GROUP 3

Problem 1. In how many ways can we place a digit instead of @ so that will be true?

A) 3 B) 4 C) 5

Problem 2. The unknown addend @ in the equation 3 dm = @ dm + 20 cm is:

A) 1 B) 10 C) 17

Problem 3. To what number do you add 13 to get 101?

A) 87 B) 88 C) 89

Problem 4. The number 3 is NOT the sum of:

A) 3 consecutive numbers B) 4 consecutive numbers C) 2 consecutive numbers 
Problem 5. How many even numbers is the magic square made of?

6 8 1

Hint: 0 is an even number.

A) 6 B) 7 C) 9

Problem 6. How many sheets of paper are there between the 23 and 77 pages of a book?

A) 25 B) 26 C) 55

Problem 7. From which number, if we subtract 11, we will get 89?

A) 80 B) 90 C) 100

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A) 11 B) 21 C) 29 
Задача 9. Fill in the missing number in the box.

 3 2 + 15  25 = 0

A) 18 B) 15 C) 12

Problem 10. How many are the numbers between 12 and 120, which have at least two digits of 1?

A) 10 B) 11 C) 12

Problem 11. What is the number of possible different sums that we get when we add the results 
from throwing 4 dice?

Problem 12. Find the value of

Problem 13. There were 9 pieces of paper. Some of them were cut into three parts. Altogether, 
there are now 19 pieces of paper. How many pieces were cut into three parts?

Problem 14. A textbook is opened at random. To what pages is it opened if the sum of the facing 
pages is 89?

Problem 15. What are the last 2 digits of the sum

⏟

Problem 16. How many numbers between 1 and 99 are divisible by 2 and 6? 
Problem 17. It is known that :

- Among A, B , C and D there are two excellent students;
- Among A, B and C there is one excellent student;
- Among A, C and D there is one excellent student.
How many are the excellent students?

Problem 18. How many seconds do we have to take out of 72 seconds to get 1 minute?

Problem 19. Use 1, 2, 3, 4 and 5 to form a 2-digit number and a 3-digit number. Find the largest 
sum of these two numbers.

Problem 20. How many sticks with a length of 11 cm can we cut off from a stick with a length of 1

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 B) 4 139>100; 139>110; 139>120; 139>130

2 A) 1 3 dm = 1 dm + 20 cm

3 B) 88 101 – 13 = 88

4 B) 4 3=1+2; 3=0+1+2

5 A) 6 6 8 1

0 5 10

9 2 4

6 B) 26 These are the lists of paper with page numbers (25,26),…, (75,76).

7 C) 100 100 – 11 = 89

8 B) 21 If the first 20 pencils are two of the colors, the 21st will be of the third
color.

9 B) 15 ((0+25) - 15)÷ 2 3 = 15

10 B) 11 101,110, 111, …, 119

11 21 The smallest sum is 1+1+1+1=4, …, the greatest is 6+6+6+6=24.
From 4 to 24 the possible sums are 21.

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13 5 If we cut 1 list of paper into 3 smaller lists, their number would be
8 + 3 = 11.

If we cut 2 lists of paper into 3 smaller lists, their number would be
7 + 6 = 13.

If we cut 3 lists of paper into 3 smaller lists, their number would be
6 + 9 = 15.

If we cut 4 lists of paper into 3 smaller lists, their number would be
5 + 12 = 17.

If we cut 5 lists of paper into 3 smaller lists, their number would be
4 + 15 = 19.

14 44 and 45 89 = 44 + 45

15 85 1+4+9+16+25+36+49+64+81=285

16 16 The numbers are 6, 12, 18, …, 90, 96.

17 2 If A is an excellent student, then from the second and third statement it
follows that B, C and D cannot be excellent students. Therefore 1 of
the statements is not true. A is not an excellent student.

If B is an excellent student, then C cannot be an excellent student (as 
follows from the second statement). Therefore D must be an excellent
student.

If B is not an excellent student, then C and D would be excellent

students (first statement). However in this case the third statement
could not be true.

Answer: The excellent students are B and D. There are two of them.
18 12 72 – 12 = 60 seconds = 1 minute

19 573 573= 542+31=541+32=531+42=532+41

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WINTER 2016: GROUP 3

Problem 1. What is the missing number?

A) 142 B) 258 C) 242

Problem 2. The sum of is:

A) 750 B) 730 C) 650

Problem 3. What is the missing number?

A) 6 B) 7 C) 8

Problem 4. How many of the following expressions are correct?

A) 1 B) 2 C) 3

Problem 5.

A) B) C)

Problem 6. What is the sum of the numbers in the 9th row?

1 2 3

4 5 6

7 8 9

... ... ...

A) 99 B) 78 C) other

Problem 7. There is a basket in a dark room. In the basket there are 6 yellow, 5 red and 4 green 
apples. What is the smallest possible number of apples we would need to take out, without looking
at their colour, in order to ensure that we have taken out apples from all three colours?

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Problem 8. If a person participating in an archery contest hits the target in all 3 attempts, how many 
different results can they get? Keep in mind that the result is the sum of all 3 attempts. (If the arrow
falls between 8 and 9, 9 points are counted. If it falls on the line between 9 and 10, 10 points are
counted.)

A) 5 B) 6 C) 7

Problem 9. If we add the number equal to 300 + 100 to the number equal to 200 + 400, we would 
get

A) 100 B) 1,000 C) 1,100

Problem 10. A gallery has 360 paintings. 101 of them were sold on the first day. The paintings sold 
on the second day were 31 more than those sold on the first day. How many paintings are still not
sold?

А) 132 B) 127 C) 228

Problem 11. Three friends weigh respectively 24, 30 and 42 kilograms. They want to cross a river 
by using a boat that can carry a maximum of 70 kg. At least how many times would this boat need 
to cross the river, so that all three of them would get to the opposite shore?

Problem 12. By how much is the larger sum greater than the smaller sum?

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Problem 13. In how many rectangles can we find the ant? (Keep in mind that a square is also a 
rectangle.)

Problem 14. Place the digits 1, 2, 7, 8 and 9 in the squares in such a way that after calculating, the 
result would be the greatest possible number. What is the number?

+ +

Problem 15. Boko and Tsoko went fishing with their sons. All of them caught an equal number of 
fish. How much fish did each of them catch, if they caught 9 fish in total?

Problem 16. What is the greatest possible sum of three odd one-digit numbers?

Problem 17. The numbers 1, 10, 19, 28,…, 82, 91 have been written down according to the
following rule: we get each following number by adding 9 to the preceding number, until we reach
91. How many numbers have been written down?

(Hint: 10 = 1 + 1 9; 19 = 1 + 2 9; 28 = 1 + 3 9, … )

Problem 18. I chose a number. I subtracted 555 from it and got 166 as a result. What was the 
number I chose?

Problem 19. What is the smallest three-digit number with 18 as the sum of its digits?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 В ? – 58 = 200; 
? = 200 + 58 = 258

2 А 247 + 178 + 325 = 425 + 325 = 750

3 А 5 = 30;

= 6

4 С All three expressions are correct.
5 С The value of the expression is 6.

The possible answers are respectively 24, 24 and 6.

6 В The numbers in the 9th row are 25, 26, and 27 (read from left to right).
Their sum is 78.

7 B In the worst case scenario we would first take out all 6 yellow apples, after that
we would take out all 5 red apples, after which we would take out 1 green
apple. We would then have taken out apples from all three colours.

6 + 5 + 1 = 12

8 С The smallest result is 24, and the greatest is 30. There are 7 possible results in
total:

24 = 8 + 8 + 8; 25 = 8 + 8 + 9; 26 = 8 + 9 + 9 = 8 + 8 + 10; 27 = 9 + 9 + 9 = 8
+ 9 + 10; 28 = 9 + 9 + 10 = 8 + 10 + 10; 29 = 9 + 10 + 10; 30 = 10 + 10 + 10
9 B (300 + 100) + (200 + 400) = 400 + 600 = 1000

10 В The paintings sold on the second day were 101 + 31 = 132. The paintings sold
on the first and second day together were 233. The paintings remaining unsold
in the gallery are 360 – 233 = 127.

11 3 Let C denotes the heaviest of the three friends, A - the lightest one, and B - the 
third one.

It would be impossible for all three of them to cross the river in one go,

because 24 + 30 + 42 = 96 > 70.

Therefore the boat would have to return at least once, and the smallest possible
number of river crossings would be 3.

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to the opposite shore:

C stays on one of the shores, while A and B cross over to the opposite shore.
A crosses back to the initial shore.

A and C now cross to the opposite shore together.

12 219

The first sum is 136, and the second sum is 355.
355 – 136 = 219

13 6 The ant can be found in 1 square 1 1, in 2 rectangles 1 2, in 1 rectangle 1 3;
in 1 square 2 2, in 1 rectangle 3 2.

14 12 7 + 8 + 9 – 12 = 12

15 3 or 1 If we assume that the problem speaks of four people – two fathers and two
sons, then the result would be impossible, because 9 is not divisible by 4.
Therefore the problem must speak of three people: a grandfather, his son, and
his grandson, or of 9 people: two fathers and seven sons.

16 27 9 + 9 + 9 = 27

17 11 We get the second number by adding 9 once (1 + 1 9) to 1. We would get 91
by adding the number 9 ten times to the number 1 (1 + 10 9). Therefore the
numbers that have been written down are 11.

18 721 555 + 166 = 721

19 189 The number is presented as 1xy, where x + y = 17. 17 is presented in two ways: 
as the sum of the two numbers 9 and 8, or 8 and 9. The numbers of the type
1xy are two: 189 and 198. The number we are looking for is 189.

20 50 9 + 91 + 18 + 82 + 27 + 73 + 36 + 64 + 45 + 55 = (9 + 91) + (18 + 82) + (27 +
73) + (36 + 64) + (45 + 55) = 100 + 100 + 100 + 100 + 100 = 500

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SPRING 2016: GROUP 3

Problem 1. If then = ?

A) 54 B) 48 C) 60

Problem 2. 1,000 – (12 + 23 + 34 + 45 + 55 + 66 + 77 + 88) = ?

A) 400 B) 500 C) other

Problem 3. One kg of dried mushrooms is derived from 12 kg of fresh mushrooms. How many kg 
of fresh mushrooms would you need to get 6 kg of dried mushrooms?

A) 2 B) 18 C) 72

Problem 4. Two ants are moving towards each other. One of them travelled a distance of 176 cm, 
and the other travelled 80 mm more than the first. What is the length that both ants travelled in 
total?

A) 36 dm B) 260 cm C) 402 mm

Problem 5. The product of 4 natural numbers is 72. The sum of these numbers is 15 and neither of 
them is 2. Which is the greatest among these numbers?

A) 6 B) 8 C) 9

Problem 6. We have 5 identical chocolate bars, each consisting of 28 pieces. We have to divide 
them equally between 7 children. What is the minimum number of times we need to break each
chocolate bar in order to do this?

A) 6 B) 7 C) 8

Problem 7. The sum of the three-digit numbers ̅̅̅̅̅, ̅̅̅̅̅̅ and ̅̅̅̅̅ is 1010. ( and represent
missing numbers). In this case, what is the three-digit number ̅̅̅̅̅̅

A) 382 B) 371 C) 473

Problem 8. A book has been numbered as follows: the first pair of pages has been numbered as 1 
and 2; the second pair as 3 and 4, and so on, until the last pair of pages, which has been numbered
as 127 and 128. If I open the book at a random place, what is a possible product of the numbers of

the two pages that I’ve opened the book at?

A) 90 B) 72 C) 56

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A) 4 B) 6 C) 8

Problem 10. How many digits are used to write down the first 100 odd numbers?

A) 250 B) 245 C) 200

Problem 11. Four children met together: Adam, Bobby, Charley and Daniel. Adam shook hands 
with 3 of these children, Bobby shook hands with 2, and Charley shook hands with 1. How many of
the children’s hands did David shake?

Problem 12. I solve 6 problems a day, and my brother solves three times less. Together we solved 
72 problems. How many days did it take us to do this?

Problem 13. The product of a few different one-digit numbers is a number that has a 5 in the ones 
place. How many even numbers are there among the multipliers?

Problem 14. Between each two neighbouring digits of the number 2016, I placed either 2 addition 
signs and 1 multiplication sign, or 2 multiplication signs and 1 addition sign.

Example: or .

How many different numbers will I get after calculating all such expressions?

Problem 15. Annie has a magical necklace. Each bead of the necklace is numbered (1, 2, 3, 4 and 
so on). If between the beads numbered as 5 and 15 there is the same number of beads, what is the
total number of beads on Annie’s necklace?

Problem 16. In Rose’s garden there are 232 roses which are not in bloom yet and 168 which are
blooming. Every day 4 new roses bloom and the ones that are already blooming do not fade. How
many days will it take for the blossoming and non-blossoming roses to be an equal number?

Problem 17. A vessel, when full of water, weighs 20 kg, and when half full it weighs as much as 3 
empty vessels. How many kilograms does this vessel weigh when it is empty?

Problem 18. A square has a side length of 1 cm. On each of its sides (on the outside) has been built 
another square with a side length of 1 cm. After that, on each of the sides of the new figure, another 
square with a side length of 1 cm has been built. How many cm is the perimeter of the final figure? 
Problem 19. What is the maximum possible number of different odd three-digit numbers that we 
can add and receive a three-digit number as a result?

Problem 20. The expression we are going to use for this problem is

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 A ⇒ =54
2 C 1000 – (12 + 23 + 34 + 45 + 55 + 66 + 77 + 88) =

1000 – (100 + 100 + 100 + 100) = 1000 – 400 = 600

3 C One kg of dried mushrooms can be derived from 12 kg of fresh 
mushrooms. In order to get 6 kg of dried mushrooms we would need 
6 12 = 72 kg fresh mushrooms.

4 A One of the ants travelled a distance of 176 cm, and the other traveled 
176 + 8 = 184 cm. The distance that both ants travelled in total is equal to 
176 + 184 = 360 cm = 36 dm.

5 В 15 = 8 + 3 + 3 + 1 = 6 + 6 + 2 + 1 = 9 + 2 + 2 + 2
Therefore the number we are looking for is 8.

6 А The number of all pieces of all five chocolates is 5 28 = 140.
Therefore each child should get 140 7 = 20 pieces.

By breaking one chocolate, we can get 20 pieces and have 8 left.
In this way we can give 20 pieces to 5 children, however there would be 2
more children and 5 more parts, each consisting of 8 pieces, left.

We can give 2 parts with 8 pieces each to each of the two children, and the
fifth part, which consists of 8 pieces, we can divide in 2 parts of 4 pieces.
The number of times we would need to break the chocolates is 5 + 1 = 6.
7 B ̅̅̅̅̅ + ̅̅̅̅̅̅ + ̅̅̅̅̅ = 1010

⇒
⇒

⇒ Therefore ̅̅̅̅̅̅

8 B If I open the book, there will be two pages, both numbered. The smaller
number will be even, and the greater will be odd. The numbers will be
consecutive.

90 = 9 10; 72 = 8 9 and 56 = 7 8, therefore I have opened the book
at the pages numbered 8 and 9.

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grandmothers, who are also mothers, they have one daughter each, i.e. 2
daughters, each of whom is also a mother to 1 granddaughter – 2
granddaughters, who are also daughters.

The two granddaughters are also 2 daughters.

There are now 2 daughers left, who are also 2 mothers.
There are now 2 mothers left, who are also 2 grandmothers.

10 B Among the first 100 odd numbers there are 5 one-digit numbers, 45
two-digit numbers and 50 three-two-digit numbers.

Therefore the number of digits which have been used to write them down
is 5 1 + 45 2 + 50 3 = 245.

11 2 A B C D

A + + +

B + – +

C + – –

D + + –

If we add the number of hand shakes, the number must be divisible by 2,
because each hand shake is counted twice.

In this case the number of hand shakes is 6 + x.

We can mark the number of David’s handshakes with x. The number x 
can NOT be greater than 3.

6 + x can be divided by 2 only if x is either 0 or 2.

However, x is not 0, because Adam shook hands with all the children. 
Therefore x = 2. David shook hands with 2 children.

12 9 I solve 6 problems a day, and my brother solves 2. Together we solved 8
problems in total. It would take us 9 days to solve 72 problems.

13 0 The number 5 is among the multipliers. If there is at least one even
number, then the product would be divisible both by 2, and by 5, i.e. by
10, it would have a ones digit of 0. Among the multipliers there are no
even numbers.

14 4 Here are all the different options:

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The result consists of 4 numbers: 2, 6, 7 and 8.

15 20 The beads with numbers from 6 to 14 are situated between the beads with
numbers from 5 to 15. The beads are 9 in total. The beads on the opposite
side are also 9. If we also note the 2 beads numbered 5 and 15 we will find
that the beads on Annie’s necklace are 2 9 + 2 = 20.

16 8 The roses which are blooming and the roses which are not yet in bloom
are 400 in total. The number of the roses in bloom needs to be increased
by 32 roses. This will happen in 32 3 = 8 days.

17 4 The weight of the water in a half-full vessel is equal to two empty vessels.
The weight of the water in a full vessel weighs as much as 4 empty
vessels. The weight of the vessel plus the water inside it is equal to 5
empty vessels. Therefore one empty vessel would be equal to 20 5=4 kg.
18 20 There are 4 squares with 3 sides each and 4 squares with 2 sides each

which form the final figure. Therefore we get that 4 3 + 4 2 = 20 cm. 
19 9 101 + 103 + 105 + 107 + 109 + 111 + 113 + 115 + 117 = 981;

101 + 103 + 105 + 107 + 109 + 111 + 113 + 115 + 117 + 119 = 1100.
20 2 In order for the value of the expression to be increased by 1, the following

needs to be true:

The first number in the expression 3+12-10 needs to be exchanged with 4.
Then the initial value would be increased by 1.

2= 4 would be possible if we exchange 6 for 8.
6 = 4 is not possible.

The second number in the expression 3+12-10 needs to be exchanged with
13. Then the initial value of the expression would be increased by 1.

3= 13 is not possible.
4 =13 is also not possible.

The third number in the expression 3+12-10 needs to be exchanged with 9.
Then the initial value would be increased by 1.

1 = 9, if we exchange 10 for 9.

10=9 is not possible.

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FINAL 2016: GROUP 3

Problem 1. If ̅̅̅̅̅ then Δ = ?

A) 10 B) 12 C) 14

Problem 2. 1000 – (5 + 15 + 25 + 35 + 45 + 55 + 65 + 75 + 85 + 95) = ?

A) 400 B) 500 C) 600

Problem 3. If we have 1 kg of fresh mushrooms and we dry them, we would get 100 g of dried 
mushrooms. How much fresh mushrooms do we need in order to derive 2 kg of dried mushrooms? 
A) 10 kg fresh mushrooms B) 20 kg fresh mushrooms C) 30 kg fresh mushrooms

Problem 4. The segment AB is 1 km long and has been divided into 1000 equal parts by a number 
of points. The points have been numbered, with A being the first point and B being the last point. 
Point C is found at an equal distance between point 101 and point 203. What is the distance (in 
meters) from point A to point C?

A) 150 B) 151 C) 152

Problem 5. Iva arrived at the bus stop and looked at her watch, which showed the time to be 
08:01h, which meant that she was 2 minutes late for her bus. What she did not know was that her
watch was running 5 minutes ahead. If the bus came 1 minute late, how many minutes did Iva have
to wait at the bus stop?

A) 4 B) 5 C) more than 5

Problem 6. Now many times at least would we need to break 6 chocolates, in order to divide them 
equally between 4 children? Each chocolate is made up of 28 pieces.

A) 2 B) 4 C) 7

Problem 7. A book is numbered as follows: the pages on the first sheet are numbered as 1 and 2, 
the pages on the second sheet are numbered as 3 and 4, and so on, until the last sheet, where the
pages have been numbered as 47 and 48.

If I were to rip off 3 consecutive sheets and then add the 6 numbers with which the pages of the
sheets have been numbered, which of the following sums would I get?

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Problem 8. If we exchange the identical letters with identical numbers, and the different letters with 
different numbers, then what would the greatest possible value of the following expression be equal
to?

A) 157 B) 156 C) 158

Problem 9. Adam wrote down 35 numbers. The first number he wrote is 7, and each next number 
is twice as big as the preceding one. How many of the numbers he wrote down are greater than
224?

A) 30 B) 29 C) 28

Problem 10. On the figure below you can see that in the middle there is a square with a side of 
1cm. On each of its sides there is another square, each with sides of 1cm. On each of the sides of
the newly formed figure, there is one extra square with a side of 1cm. What is the minimum number
of squares that must be erased in order for only 15 squares to remain on the figure?

A) 1 B) 2 C) more than 2

Problem 11. If the first day of the year is a Monday, what would the last day of the same year be? 
Problem 12. Alex and Boris each have 3 coins of 1, 2 and 5 cents. Boris used 7 of those coins to 
add up the smallest possible sum and Alex used 7 of those coins to add up the greatest possible
sum. By how much is Boris’ sum smaller than that of Alex?

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Problem 14. You are given the following expression: – . Exchange one of
the numbers in the expression with a different number, so that the initial value of the expression
would be increased by 1. In how many ways can we do this?

Problem 15. How many times is the number hidden under the first shell smaller than the number 
hidden under the second shell?

1, , 2, 6, 24, , 720, 5 040

Problem 16. At least how many of the numbers do we need to change in order for the product of 
the numbers along the diagonals, rows and columns to be the same?

1 4 8

16 4 1

2 4 8

Problem 17. Each of the 10 digits has been used once to write down 5 two-digit numbers with the 
greatest possible sum. What is the sum?

Problem 18. At a football game, the winner earns 3 points and the loser earns 0 points. If the match

is drawn, both teams get 1 point each. After having played 7 games, a team had earned 11 points.
What is the possible number of losses that the team had?

Problem 19. A number is perfect when the sum of its divisors (except the number itself) equals the 
given number. The number 6 is called perfect because it is equal to the sum of

1 + 2 + 3, where 1, 2 and 3 are all its divisors, except the number 6. The next perfect number is an
even number greater than 24 and smaller than 30. What is the number?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 В 332 4=1328 ⇒ Δ=12

2 B

1000 – (5 + 15 + 25 + 35 + 45 + 55 + 65 + 75 + 85 + 95) =
= 1000 – 500 = 500

3 B

2 kg = 2000 grams = 20 100 grams, which means that we would need 
20 kg of mushrooms.

4 В

Point C, which is found at an equal distance between point 101 and point 
203, is point 152. The distance from the first point to the 152nd point is
151 meters.

5 А

The watch shows that the time is 8:01h. She is 2 minutes late, which
means that she was supposed to arrive at 7:59h, according to her watch.
This means that she came 3 minutes early, because the bus (according to
her watch) was supposed to arrive at 8:04h. The bus is running 1 minute
late. Therefore it would arrive at 8:05h. Iva had to wait for 4 minutes.

6 А

Each child will receive 1 chocolate plus extra 14 pieces of the remaining
chocolates. In this case we would only need to break the chocolates twice.

7 C

Everything follows from:

⏟

⏟

45.

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9 В

The numbers are 7, 14, 28, 56, 112, 224, 448,...
5 of them are smaller than 224, 1 is equal to 224,
35 – 6 = 29 are greater than 224.

10 А

The 1 1 squares are 13, the 2 2 squares are 4, and there is one 3 3
square. There are 18 squares in total.

If we were to remove one 1 1 square, the 1 1 squares would be 12, the
2 2 squares would be 3, and the 3 3 squares would be 0, therefore the
total number of squares would be 15.

This can be done in 4 ways, by removing 1 square from one of the 4
angles of the the 3 3 square.

11

Monday 
or

Tuesday

The calendar year has 365 days, or 366 days on a leap year. When
dividing 364 and 365 by 7, the remainders are 0 and 1. Therefore the last
day of the year would either be a Monday or a Tuesday.

12 8

The smallest possible sum is 3 1 + 3 2 + 1 5 = 14, and the greatest
possible sum is 3 5 + 3 2 + 1 1 = 22. The difference is 22 – 14 = 8.

13 А or D

First we need to arrange 4 people as follows: DCEA or AECD 
In this case B may be situated as follows:

ВDСЕA 
DСЕAВ 
АECDВ 
ВАECD

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14 3

– ;
– ;
–
We can do that in 3 ways.

15 120

The rule is as follows: multiply the first number by 1, and get the second;
multiply the second number by 2, and get the third, etc.

1; 1 1 = 1; 1 2 = 2; 2 3 = 6; 6 4 = 24, 24 5 = 120; 120 6 =
720; ...

The number hidden under the first shell is 1, and the number hidden under
the second shell is 120. 120 1 = 120.

16 1

If we exchange the number in the first row, first column, with a 2, we
would get a magical square.

2 4 8
16 4 1
2 4 8

17 360 90+81+72+63+54=360.

18 0 or 2

If a team were to win all 7 games, they would earn 21 points. This team
however earned 11 points, which means that they lost 10 points from the
maximum score.

This can be done through:
5 drawn games and 0 losses,
2 drawn games and 2 losses;
Answer: 0 or 2 losses.

19 28 The number is 28. 28 = 1 + 2 + 4 + 7 + 14.

20 450

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TEAM COMPETITION – NESSEBAR, BULGARIA 
MATHEMATICAL RELAY RACE

The answers to each problem are hidden behind the symbols @, #, &, § and * and are used in 
solving the following problem. Each team, consisting of three students of the same age group, must
solve the problems in 45 minutes and then fill a common answer sheet.

GROUP 3

Problem 1. If ⏟

a n s , find @.

Problem 2. Our rabbit now has less than @ male and female bunnies. Each male bunny has as 
many sisters as brothers, and each female bunny has half as many sisters as brothers. If the number
of bunnies our rabbit has is #, then find #.

Problem 3. Find the smallest even three-digit number &, if it is known that & 5 is divisible by #.

Problem 4. The number #+2 is presented as the product of 4 consecutive odd numbers with a sum 
of §. Find §.

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 @ = 10

⏟

40 ⇒@ = 10

⏟

a n s

40 ⇒@ = 10

2 # =7

The condition says that each male bunny has as many sisters, as brothers.
Therefore the number of baby bunnies our rabbit has is 3, 5, 7 or 9.
We can check each possible answer:

If the correct answer is 3, then each male bunny would have 1 brother
and 1 sister. However the only female bunny in this case would not have
“half as many sisters as brothers”.

If the correct answer is 5, then each male bunny would have 2 brothers
and 2 sisters. However in this case each female bunny would have 1 sister
and 3 brothers, therefore it would not have not have “half as many sisters
as brothers”.

If the correct answer is 7, then each male bunny would have 3 brothers
and 3 sisters. In this case each female bunny would have 2 sisters and 4
brothers, which means that the condition is satisfied – each female bunny
would have “half as many sisters as brothers”.

If the correct number is 9, then each male bunny would have 4 brothers
and 4 sisters. In this case each female bunny would have 3 sisters and 5
brothers, therefore it would not have “half as many sisters as brothers”.

3 & =103

If we carry out a check, we will find that among the differences 100 5,
101 5, 102 5, 103 5, 104 5, 105 5,...., the first one that is divisible
by #, i.e. by 7, is 103 5. The number we are looking for is 103.

& = 103.

4 § =16 The number 103 + 2 = 105 = 1 3 5 7.
The sum of these multipliers is 16.

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MATHEMATICS WITHOUT BORDERS 
2015-2016

AUTUMN 2015: GROUP 4 
Problem 1. What is the largest 4-digit number with 0 as a units digit?

A) 9,909 B) 9,990 C) 9,099

Problem 2. How many are all the possible digits that can be placed instead of @, so that 
would be true?

A) 0 B) 8 C) 9

Problem 3. I thought of a number. I added it to 222 and got 1,000. The number I thought of is:

A) 1,222 B) 888 C) 778

Problem 4. How many are the 5-digit numbers that do NOT have 9 as a units digit?

A) 810 B) 8,100 C) 81,000

Problem 5. How many sheets of paper are there between the and the pages of a book?

A) 99 B) 98 C) 48

Задача 6. Fill in the missing number in the box.

 20 + 15  2,015 = 0

A) 400 B) 4,000 C) 40,000

Problem 7. Find the value of

A) 2,025 B) 2,020 C) 2,015

Problem 8. There were 1,001 pieces of paper. Some of them were cut into three parts. Altogether, 
there are now 2,015 pieces of paper. How many pieces were cut into three parts?

A) 507 B) 494 C) 1,014

Problem 9. If the difference is 9,999 and the subtrahend is 1, the minuend is:

A) 9,998 B) 10,000 C) 1,000

Problem 10. A student only marked the odd numbered pages of his notebook by using only odd 
numbers such as 1, 3, 5 etc. He used 93 digits. How many pages does the notebook have?

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Problem 11. I chose a number and added it to 1. Then I multiplied the resulting sum by 2. After that I 
divided the resulting product by 3. What is the number I chose, if the quotient is 4, and the remainder
is 2?

Problem 12. The three following actions have been applied to the number 2 in a random order: 
- multiplying by 2

- dividing by 2
- adding 2

How many possible results are there?

Problem 13. There are less than 100 apples in a basket. These apples can be divided equally between 
2, 3, or 5 children. These apples can NOT be divided equally between 7 children - one more apple
would be needed to do that. What is the number of apples in the basket?

Problem 14. In ̅̅̅̅ ̅̅̅̅̅̅ each letter corresponds to a digit. Identical letters correspond to
identical digits and different letters correspond to different numbers. What is the greatest possible
number that corresponds to ̅̅̅̅̅̅̅̅̅̅?

Problem 15. How many missing addends are there in the expression 
?

Problem 16. Which of the following is the smallest number: 62 345, 523 420 and 432 100?

Problem 17. How many pairs of integers which product is 63 can be selected from 0 to 99?

Problem 18. What is the next number in the sequence of numbers? 
2, 11, 20, 101, 110, 200, 1001, 1010, 1100, 2000, 10 001, ...

Problem 19. In the number A the positions of the tens and hundreds were exchanged and the 
resulting number was 1,234. What is the number A?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 В) 9,990 *,**0 ⇒9,990

2 С) 9 2015 < 2115, 2015 < 2215,…, 2015 < 2915

3 C) 778 1000 222 = 778

4 C) 81,000 The numbers are

5 С) 48 These are the sheets s of paper with page numbers (5; 6), …, (99; 100).
Their total number is 48.

6 C) 40,000

7 А) 2,025

8 А) 507 The number of pieces increases by a number that is the doubled
number of the pieces that have been cut. The number of pieces has
increased by 1014. Therefore the number of pieces that have been cut
is 507.

Another way to find the answer is to check the given answers

9 B) 10,000 9,999+1=10,000

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11 6 Let us solve the problem starting at the end: the number which we
divided by 3 is 4 3 + 2 = 14.

The number that we multiplied by 2 is 14 2 = 7.
The number that we added to 1 is 7 – 1 = 6.

12 3 If we mark the actions with M, D and A, there are 6 ways for their
sequential use: MDS, MSD, DMS, DSM, SMD, SDM.

For each of these ways we obtain the following results:
4, 3, 4, 6, 4, 4.

Thus the number of different results is 3: 3, 4 and 6.

13 90 The numbers divisible by 5 and by 2 with a remainder of 0 are: 10, 20,
30, 40, 50, 60, 70, 80, 90. The ones divisible by 3 are 30, 60 and 90.
From the numbers 30+1, 60+1 and 90+1 only 91 is divisible by 7.
There are 90 apples in the basket.

14 98,107 The sum ̅̅̅̅ is at most ̅̅̅̅
Therefore ̅̅̅̅̅̅̅̅̅̅

15 23 The addends are the one digit (except 0) and two digit numbers,
divisible by 3. The first number is 3, and the 33rd number is 99. The
numbers that have been skipped are 33 – (7 + 3) = 23.

16 62,345 62,345 < 432,100 < 523,420

17 3 The numbers are 1 and 63; 3 and 21; 7 and 9.

18 10,010 These are the numbers with 2 as the sum of their digits. They have
been arranged from smallest to biggest. The next number is 10,010.

19 1,324 1,234 ⇒1,324

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WINTER 2016: GROUP 4 
Problem 1. What is the missing number?

A) 1,000 B) 1,010 C) 990

Problem 2. Which three numbers are smaller than 30,020?

A) 30,019; 30,020; 30,021 B) 30,001; 30,010; 30,019 C) 30,001; 30,010; 31,000 
Problem 3. If the difference is 24,345, and the subtrahend is 6,707, the minuend is:

A) 31,052 B) 17,638 C) 17,648

Problem 4. How many of the following expressions are correct? 
165 + 561 = 727
264 5 – 2 = 264 3
90,000 10 < 10,000

A) 1 B) 2 C) 3

Problem 5. We have an apple, a pear, an orange and a lemon.

We must distribute them among two children.
In how many different ways can we distribute the fruit, so that each child would get 2 fruit?

A) 8 B) 6 C) 4

Problem 6. What is the hundreds digit of the smallest 5-digit number, that has a sum of its digits 
equal to 25?

A) 6 B) 9 C) 1

Problem 7. Which of the following numbers has 3 as a digit of hundreds and 1 as a digit of 
thousands?

A) 1,313 B) 3,311 C) 3,113

Problem 8. Last year 33 white, red and yellow tullips blossomed at the same time in Maya’s garden.
The white and red tullips together were 19, and the red and yellow tullips together were 18. Which
tullips were of the greatest number?

A) red B) yellow C) white

Problem 9. The four-digit numbers smaller than 2,015 are:

A) 2,014 B) 1,015 C) 1,016

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*1
+*96
92*

2016

A) 9 B) 8 C) 7

Problem 11. The natural numbers А, B, C and D are such that A × B = 2, B × C = 6 and C × D = 3. 
What is the number D?

Problem 12. Three friends weigh respectively 24, 30 and 42 kilograms. They want to cross a river by 
using a boat that can carry a maximum of 70 kg. At least how many times would this boat need to
cross the river, so that all three of them would get to the opposite shore?

Problem 13. The numbers 1,001; 1,008; 1,015; 1,022; …; 2,016 are recorded according the following 
rule: we get each following number by adding 7 to the preceding number, until we reach the number
2,016. How many numbers are there in total?

(Hint: 1,008 = 1,001 + 1 7; 1,015 = 1,001 + 2 7; 1,022 = 1,001 + 3 7, ….. ) 
Problem 14. In how many rectangles do we find just one ant?

Problem 15. Place the numbers 1, 2, 3 and 4 in the squares in a way that would result in the greatest

possible product. What is the product?

Problem 16. I have 5 baskets, each of which has 55 apples inside. If I move the apples to 11 baskets, 
and each basket has the same number of apples, how many apples would there be in each basket?
Problem 17. What is the digit of ones of the product of all odd one-digit numbers?

Problem 18. When A is divided by 5 the remainder is 2. What is the remainder when 3A is divided 
by 5?

Problem 19. What is the missing number?  2 = 330 5 + 330

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 А ? + 110 = 1,110, i.e. ? = 1,000

2 В 30,001; 30,010; 30,019

3 А ? – 6,707 = 24,345
24,345 + 6,707 = 31,052

4 А 165 + 561 = 726, i.e. the first expression is wrong.

264 5 – 2 = 1318; 264 3 = 792, i.e. the second expression is wrong.
90,000 10 = 9,000 < 10 000, i.e. the third expression is correct.

Of the three expressions, only the third one is correct.

5 B There are 6 possibilities. If we number the fruit as 1, 2, 3 and 4, then they

would be distributed as follows:

First child Second child

1, 2 3, 4

1, 3 2, 4

1, 4 2, 3

2, 3 1, 4

2, 4 1, 3

3, 4 1, 2

6 А 25 = 9 + 9 + 6 + 0 + 1, therefore the smallest five-digit number with a sum of
its digits equal to 25, is 10699.

7 А In the number 1,313 the digit of ones is 3, the digit of tens is 1, the digit of
hundreds is 3, and the digit of thousands is 1.

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plus the number of red tullips. Therefore, the number of red tullips is

19 + 18 – 33 = 4, the number of white tullips is 15, and the number of yellow
tullips is 14. The white tullips are of the greatest number.

9 В The numbers from 1 to 2,014 are 2,014. Among those numbers, the numbers
from 1 to 999 are not four-digit numbers. Therefore the number we are looking
for is 2,014 – 999 = 1,015.

10 А We can solve the problem by carrying a check by using the possible answers.
We can start as follows:

1 + 6 + * = …6, if * = 9.

11 1 If A × B = 2 ⇒ A = 1 or A = 2. If A = 1 ⇒ B = 2 ⇒ C = 3 ⇒ D = 1.
If A = 2 ⇒ B = 1 ⇒ C = 6 ⇒ D could not be a natural number such that
6 × D = 3.

Hence D = 1.

12 3 Let C denotes the heaviest of the three friends, A - the lightest one, and B - the 
third one.

It would be impossible for all three of them to cross the river in one go,
because 24 + 30 + 42 = 96 > 70.

Therefore the boat would have to return at least once, and the smallest possible
number of river crossings would be 3.

Following is an example of a way in which all three friends can cross the river
to the opposite shore:

C stays on one of the shores, while A and B cross over to the opposite shore.

A crosses back to the initial shore.

A and C now cross to the opposite shore together.

13 146 2,016 = 1,001 +  7. Therefore  7 = 1,015, i.e.  = 145.

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АA1В, ВA1, А2, ВA2, ВA2С, A2C each.

15 252 All possible products are:

1 2 43 = 86; 1 2 34 = 68;
1 3 24 = 72; 1 3 42 = 126;
1 4 23 = 92; 1 4 32 = 128;
2 3 14 = 84; 2 3 41 = 246;
2 4 13 = 104; 2 4 31 = 248;

3 4 12 = 144; 3 4 21 = 252 - the greatest possible product.

16 25 The number of apples is 5 55 = 275. From 275 11 = 25, it follows that
there are 25 apples in each basket.

17 5 The number 5 is one of the multipliers, and none of the multipliers is an even
number. Therefore the product must end in 5.

18 1 First way: The numbers that when divided by 5 leave a remainder of 2, are: 2,
7, 12, 17, 22,...

The tripled numbers are: 6, 21, 36, 51, 66, ...
When divided by 5, they all leave a remainder of 1.

Second way: Since Dividend = Divisor Quotient + Remainder, we could 
present the numbers that when divided by 5 leave a remainder of 2 as follows:
5 Quotient + 2.

The tripled numbers could be presented as follows: 15 Quotient + 6.

Therefore the remainder we are looking for would be equal to the remainder of
dividing 6 by 5, i.e. the remainder = 1.

19 198  2 = 330 5 + 330, therefore  2 = 396, i.e.  = 198

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SPRING 2016: GROUP 4 
Problem 1. If  then  =:

A) 5 B) 6 C) 7

Problem 2. –

A) 2,016 B) 4,032 C) other

Problem 3. A hippo eats 200 kg of grass every day, which is three times less than the kilograms of 
grass that an elephant eats in a week. How many days would it take for a hippo and an elephant to eat
2 tons of grass?

A) 7 B) 8 C) 9

Problem 4. What is the natural number that is greater than the number which is greater by 9 than 
99,979 and smaller than the number which is 9 times greater than 11,110?

A) 99,988 B) 99,989 C) 99,990

Problem 5. The product of 4 natural numbers is 72. The sum of these numbers is 15 and neither of 
them is 2. In this case, which is the greatest among these numbers?

A) 6 B) 8 C) 9

Problem 6. We have 5 identical chocolate bars, each consisting of 28 pieces. We have to divide them 
equally between 7 children. What is the minimum number of times we need to break each chocolate
bar in order to do this?

A) 6 B) 7 C) 8

Problem 7. How many different three-digit numbers ̅̅̅̅̅ can you get as a product of the number 29
and a two-digit number?

A) 2 B) 3 C) 4

Задача 8. A book has been numbered as follows: the pages of the first sheet have been numbered as 
1 and 2; the pages of the second - as 3 and 4, and so on, until the pages of the last sheet, which have
been numbered as 227 and 228. If I open the book at a random place, what is a possible product of the
numbers of the two pages that I have opened the book at?

A) 9,900 B) 10,100 C) 90

Problem 9. In the same room we have 2 grandmothers, 4 mothers, 4 daughters and 2 granddaughters. 
What's the smallest number of people that there could be in the room?

A) 4 B) 6 C) 8

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1, 12, 123,1234, ..., 12345678 and 123456789.

A) 0 B) 6 C) 5

Problem 11. Four children met together: Adam, Bobby, Charley and Daniel. Adam shook hands with

3 of these children, Bobby shook hands with 2, and Charley shook hands with 1. How many of the
children’s hands did David shake?

Problem 12. A three-digit number consists of the digits 1, 2 and 3. The digit 1 is not the digit of 
hundreds, and the digit 3 is not next to the digit 2. What is the number?

Problem 13. The product of a few different one-digit numbers is a number that is divisible by 10 (the 
remainder is 0), but is not divisible by 20 (the remainder is not 0). Which even numbers could be
among the multipliers?

Problem 14. I placed 1 addition sign and 1 multiplication sign between the digits of the number 
2016. Example: or How many different numbers will I get after calculating
the expressions?

Problem 15. Annie has a magical necklace. Each bead of the necklace is numbered (1, 2, 3, 4 and so 
on). If between the beads numbered as 5 and 15 there is the same number of beads, what is the total
number of beads on Annie’s necklace?

Problem 16. In the garden of Rose there are 1,232 roses not yet in bloom and 1,168 that are 
blooming. Every day 4 new roses bloom and the ones that are already blooming do not fade. How
many days will it take for the blossoming and non-blossoming roses to be an equal number?

Problem 17. A container, when full of water, weighs 994 kg and when half full, it weighs as much as 
4 empty containers. How many kilograms does this container weigh when it is empty?

Problem 18. The natural numbers from 10 to 30 have been written on different cards (one on each 
card). What is the smallest number of cards that we would need to take without looking in order to
make sure that there would be at least 2 numbers divisible by 3?

Problem 19. What is the greatest possible number of different three-digit numbers that we can add

and get a three-digit number as a result?

Problem 20. The expression we are going to use for this problem is .

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Problem Answer Solution

1 A  ⇒  ⇒  = 5 
2 В –

3 А In one week a hippo can eat 1400 kg of grass, and an elephant can eat 600
kg of grass. The amount they can eat in total is 2000 kg = 2 tons.

4 B The number which is greater than 99,979 by 9 is 99,988, and the number
which is 9 times greater than 11,110 is 99,990. The number we are
looking for is 99,989.

5 В 15 = 8 + 3 + 3 + 1 = 6 + 6 + 2 + 1 = 9 + 2 + 2 + 2
Therefore the number we are looking for is 8.

6 А The number of all pieces of all five chocolates is 5 28=140.
Therefore each child should get 140 7 = 20 pieces.

By breaking one chocolate, we can get 20 pieces and have 8 left.

In this way we can give 20 pieces to 5 children, however there would be 2
more children and 5 more parts, each consisting of 8 pieces, left.

7 B 29 17=493 <5**, 29 18=522, 29 19=551, 29 20=580,

29 21=609>5**, therefore there are 3 options.

8 B If I open the book, there will be two pages, both numbered. The smaller
number will be even, and the greater will be odd. The numbers will be
consecutive.

9,900 = 99 100, 10,100 = 101 100, 90 = 9 10, therefore I have
opened the book at the pages numbered as 100 and 101. A possible
product is 10,100.

9 B In order for one of the women to be a grandmother, she would need to
have a daughter, and a granddaughter. Therefore if there are two

grandmothers, who are also mothers, they have one daughter each, i.e. 2
daughters, each of whom is also a mother to 1 granddaughter – 2

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There are now 2 daughters left, who are also 2 mothers.
There are now 2 mothers left, who are also 2 grandmothers

10 C The digit we are looking for is the last digit of the sum of the last digits,
i.e. of 1 + 2 + 3 + ... + 8 + 9 = 45.

11 2 A B C D

A + + +

B + – +

C + – –

D + + –

If we add the number of hand shakes, the number must be divisible by 2,
because each hand shake is counted twice.

In this case the number of hand shakes is 6 + x.

We can mark the number of David’s handshakes with x. The number x 
can NOT be greater than 3.

6 + x can be divided by 2 only if x is either 0 or 2.

However, x is not 0, because Adam shook hands with all the children. 
Therefore x = 2. David shook hands with 2 children.

12 312 or 
213

The number is either *1* or **1.

If the number is *1*, then the number is either 312 or 213.

If the number is **1, then the digit 3 and the digit 2 are next to each other,
which would mean that 1 is not the digit of ones.

13 2 or 6 The product of the one-digit odd numbers is a number which ends in 5. If
we multiply it by 2 or by 6 we will get a number that ends in 0, but is not
divisible by 20.

14 4 2 016 ⇒ 2 0+16=16; 2 01+6=8
20 16 ⇒2+0 16=2; 20 1+6=26

201 6 ⇒2 + 01 6 = 8; 20 + 1 6 = 26;

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by 32 roses. This will happen in 32 3=8 days.

17 142 The water in a half full container weighs as much as 3 empty containers.
The water in a full container weighs as much as 6 empty containers. The
water in a full container (the weight of the water + the actual container)
weighs as much as 7 empty containers.

Therefore an empty container would weigh 994 7 = 142 kg.

18 16 The numbers are in total. 7 of them are divisible by 3
with a remainder of 0, 7 are divisible by 3 with a remainder of 1, and 7 are
divisible by 3 with a remainder of 2.

We would need to take 7 + 7 + 2 = 16 cards, in order to make sure that we
have taken 2 cards that have such numbers on them which when divided by
3 leave a remainder of 2.

19 9 101 + 103 + 105 + 107 + 109 + 111 + 113 + 115 + 117 = 981;
101 + 103 + 105 + 107 + 109 + 111 + 113 + 115 + 117 +119 = 1100.

20 2 In order for the value of the expression to be increased by 1, the following

needs to be true:

The first number in the expression 3 + 12 – 10 needs to be exchanged with
4. Then the initial value would be increased by 1.

 2= 4 would be possible if we exchange 6 for 8. 
6 = 4 is not possible.

The second number in the expression 3 + 12 – 10 needs to be exchanged
with 13. Then the initial value of the expression would be increased by 1.
 3= 13 is not possible.

4 =13 is also not possible.

The third number in the expression 3+12-10 needs to be exchanged with 9.
Then the initial value would be increased by 1.

1  = 9, if we exchange 10 for 9.
 10=9 is not possible.

We can exchange two of the numbers in the expression, so that the initial
value would be increased by 1:

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FINAL 2016: GROUP 4

Problem 1. If ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅  , then  = ( ).

A) 1 B) 7 C) 9

Problem 2. A is a four-digit number. Its digit of tens is two times greater than its digit of ones, its 
digit of hundreds is two times greater than its digit of tens. How many A numbers with this property 
are there?

A) 1 B) 9 C) 18

Problem 3. There are several points along a straight line. A student placed a point between every two 
adjacent points. After doing this 5 times, there were now 33 points along the straight line. How many
points were originally on the straight line (before the student placed any extra points)?

A) 2 B) 3 C) 5

Problem 4. An equilateral triangle with a side length of 3 cm has been divided into 9 smaller 
equilateral triangles with a side length of 1 cm. The numbers A, B, C, 4, 5, 6, 7, 8 and 9 have been 
placed inside the smaller triangles. Now there are three equilateral triangles with sides of 2 cm on the 
diagram and the sums of the numbers inside them are equal.

What, then, is the greatest of those three numbers: A, B or C?

A) A B) B C) C

Problem 5. Iva arrived at the bus stop and looked at her watch, which showed 08:01h. It meant that 
she was 2 minutes late for her bus. What she did not know was that her watch was running 5 minutes
ahead. If the bus came 1 minute late, for how many minutes would Iva have to wait at the bus stop?

A) 4 B) 5 C) more than 5

Problem 6. The number of astronomical hours in 2016 is equal to ( ).

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Problem 7. A book is numbered as follows: the pages on the first sheet are numbered as 1 and 2, the 
pages on the second sheet are numbered as 3 and 4, and so on, until the last sheet, where the pages
have been numbered as 127 and 128. If I tear off 11 consecutive sheets and add up all the page
numbers of their 22 pages, which of the following sums is possible?

A) 255 B) 275 C) 341

Problem 8. A storage room can be filled up with either 12 chests, or with 18 boxes. There are 
currently 4 chests and 9 boxes in the room. How many more chests can fit into the room?

A) 6 B) 3 C) 2

Problem 9. A spring with a flow rate of 84 liters of water per minute provides water for three 
fountains. Four times more water reaches the second fountain than does the first, and half as much
water reaches the third fountain than does the second. How many liters per minute is the flow rate of 
the fountain which receives the greatest amount of water?

A) 56 B) 48 C) 52

Problem 10. What is the three-digit number ̅̅̅̅̅, such that ̅̅̅̅̅ < 1116 and can be presented as
the product of both 4, and of 5 consecutive natural numbers?

A) 124 B) 120 C) 100

Problem 11. Find ̅̅̅ if

2 + 24 + 246 + 2468 + 24680 + 246808 + 2468086 + 24680864 + 246808642 = ̅̅̅̅̅̅.

Problem 12. We can move from square A to square B by moving either horizontally or vertically 
from one square to another. How many different routes that go through exactly 4 squares are there?

B

A

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Problem 14. Five people (A, B, C, D and E) are waiting in a queue. C is between E and D, A is next 
to E, and B is NOT the last. Which one is the last?

Problem 15. How many three-digit numbers ̅̅̅̅̅ are there, if 17 can divide both ̅̅̅ and ̅̅̅ without a
remainder?

Problem 16. Seven numbers are arranged in a specific order and two of them are covered by shells. 
How many times is the number under the first shell smaller than the number under the second shell?

1, , 2, 6, 24, , 720

Problem 17. At least how many of the numbers do we need to change in order that the products of 
the numbers in the diagonals, rows and columns will be the same?

1 4 8

16 4 1

2 4 8

Problem 18. For a football game, the winner earns 3 points and the loser earns 0 points. If the match 
is drawn, each team gets 1 point. After 7 games, a team has earned 11 points. What is the possible
number of games that the team has lost?

Problem 19. A number is perfect when the sum of its divisors (except the number itself) equals the

given number. For example, the number 6 is called perfect because it is equal to the sum of 1 + 2 + 3,
where 1, 2 and 3 are all its divisors except the number 6 itself. The next perfect number is an even
number greater than 24 and smaller than 30. What is the number?

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Problem Answer Solution

1 C ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅  , so   would be a number digit of
ones is 1. Therefore the possibilities are either 1 or 9.

If we check both possibilities, we will find the correct answer: 9.
2 C The numbers are one of two types: either ̅̅̅̅̅̅̅ or ̅̅̅̅̅̅̅.

There are 9 possibilities for the digit of thousands for each of the two
numbers.

Therefore there are 18 numbers with such property.

3 A If the number of points is 2, then we will place 1. There are now 3 points,
we will place another 2. There are now 5 points, we place another 4. There
are now 9, we place 8. There are now 17 points, we place 16. There are
now 33 points.

4 B Let us compare the sums:

⇒ ⇒
⇒ ⇒
Answer: B.

5 А The watch showed that the time was 08:01h. She was 2 minutes late,

which means that she was supposed to arrive at 07:59h, according to her
watch. This meant that she came 3 minutes early because the bus
(according to her watch) was supposed to arrive at 8:04h. The bus was
running 1 minute late. Therefore it would arrive at 8:05h. Iva would have
to wait for 4 minutes.

6 B 2016 is a leap year, so .
7 C If I tear off sheets 1-11:

⏟

If I tear off sheets 2-12:

⏟

If I tear off sheets 3-13:

⏟

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the other half. It will take 6 chests to fill half of the space. Therefore there
is room for another 2 chests.

9 B We can carry a check using the possible answers. If the flow rate of the
second fountain is 48 liters per minute, then the flow rates of the first and 
third fountains respectively would be 12 liters per minute and 24 liters per 
minute. In this case 48 + 12 + 24 = 84.

10 B We are looking for a three-digit number smaller than 124.
The number is 120.

In fact 120 < 124 and 120 = 1 2 3 4 5 = 2 3 4 5.

11 20 246808642

24680864
2468086
246808
24680
2468
246
24
2

Add all the digits of ones of these numbers: 2 (2 + 4 + 6 + 8) = 40. 0 is
the digit of ones we are looking for and 4 need to be added to the sum of

all the digit of tens: 2 (4 + 6 + 8) + 2 + 4 = 42, 2 is the digit of tens we
are looking for.

Therefore ̅̅̅ is equal to 20.

12 10 , ,
.

c d B

b x y

a

A

13 211 The smallest sum is 3 1 + 3 2 + 3 5 + 10 = 34, and the greatest sum
is 3 50 + 3 20 + 3 10 + 1 5 = 245.

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14 А or D First we can arrange the 4 people as follows: DCEA or AECD. 
Then B may be situated as follows:

ВDСЕA 
DСЕAВ 
АECDВ 
ВАECD

Since B is not the last, A or D can be the last.

15 3 The number ̅̅̅ can be 17, 34, 51, 68 or 85. The number ̅̅̅ can be 17, 34,

51, 68 or 85. The second digit of ̅̅̅ is the first digit of ̅̅̅ Therefore the
possibilities are the following: or . We get the numbers
517, 685 and 851.

16 120 Find the pattern first: after the first number 1, the rest are equal to the
number before multiplying 1, 2, 3, 4, 5, 6 respectively, i.e.:

1; 1 1=1; 1 2 = 2; 2 3 = 6; 6 4 = 24; 24 5=120; 120 6 = 720.
Therefore the numbers under the shells are 1 and 120.

120 120.

17 1 If we change the number 1 in the first row, first column, to 2, we will get a
magic square.

18 0 or 2 11 points can be earned in the following ways:
2 wins, 5 draws and 0 defeats

3 wins, 2 draws and 2 defeats.
19 28 The number is 28.

28 = 1 + 2 + 4 + 7 + 14.

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TEAM COMPETITION – NESSEBAR, BULGARIA

MATHEMATICAL RELAY RACE

The answers to each problem are hidden behind the symbols @, #, &, § and * and are used in solving

the following problem. Each team, consisting of three students of the same age group, must solve the
problems in 45 minutes and then fill a common answer sheet.

GROUP 4

Problem 1. The maximum number of people that can attend a party is @, among whom there can not 
be two people born in the same month. Find @.

Problem 2. The sum of the digits, which are not part of the equation ̅̅̅̅̅̅̅̅̅ is #. 
Find #.

Problem 3. Use & to represent the greatest possible product of two integers with a sum of #. Find &.

Problem 4. There are 18 children in a class. Each of them has either 3 or 5 balloons. The total 
number of balloons is &. The number of children who have 5 balloons is §. Find §.

Problem 5. There are § ways to travel from point X to point A. There are * ways to travel from point

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Problem Answer Solution

1 @ = 12 If 13 people are present, there would definitely be 2 among them, who have
been born in the same month. (Dirichlet's Principle)

2 # =16 123456 12=1088 , therefore the missing digits are 7 and 9.

3 & = 64

16 = 0 + 16 = 1 + 15 = 2 + 14 = 3 + 13 = 4 + 12 = 5 + 11 = 6 + 10 = 7 + 9 =
8 + 8, therefore the possible products are 0, 15, 28, 39, 48, 55, 60, 63 and

64. The greatest is 64.

4 § = 5

If all children are carrying 3 balloons each, then 3 18 = 54. Therefore
there would be 64 54=10 balloons left. We would give those away to 5
children. In this way 13 children would be carrying 3 balloons each, and 5
children would be carrying 5 balloons each.

5 * = 60

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MATHEMATICS WITHOUT BORDERS 
2015-2016

AUTUMN 2015: GROUP 5

Problem 1. Which one is the smallest product among the following?

A) B) C) D) 
Problem 2. The number of integers from 98 to 1,000 which are divisible by 3 is:

A) 301 B) 302 C) 303 D) 304

Problem 3. If we have a 200 cm long tape and we cut off 12 dm from it, how long is the larger 
part of the two cut-offs?

A) 188 сm B) 80 сm C) 8 dm D) 12 dm

Problem 4. Peter reads 15 pages in 45 minutes. How long will it take him to read 45 pages at 
that rate?

A) 15 mim B) 1 h C) 2 h 15 min D) 1 h 15 min

Problem 5. The sum of the first 100 positive integers is 5,050. Find the sum of the first 100 
positive odd integers.

A) 10,000 B) 10,050 C) 10,100 D) 10,150

Problem 6. There were a total of 90 coins in two boxes. Ten coins were then shifted from the 
first box to the second. As a result, the number of coins in the second box was twice as much as
the number of coins in the first one. What was the number of coins in the first box before the
shift?

A) 100 B) 80 C) 60 D) 40

Problem 7. In ̅̅̅̅̅̅ ̅̅̅̅̅̅̅̅ each letter corresponds to a digit. Identical letters
correspond to identical digits and different letters correspond to different numbers. What is the
greatest possible number that corresponds to ̅̅̅̅̅̅̅̅?

A) 1007 B) 1006 C) 1005 D) 1004

Problem 8. What is the ones digit of the smallest natural number with sum of its digits equal to 
2015?

A) 9 B) 8 C) 7 D) 6

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A) 8 days B) 10 days C) 15 days D) 20 days 
Problem 10. How many squares are there in the figure below?

A) 12 B) 20 C) 22 D) 24

Problem 11. Find the value of

Problem 12. We have written down the numbers that are divisible by 5: 5, 10, 15, 20, 25, ... 
Underneath each of these numbers (in a second row) we have written down the sum of its digits.
Which place in the second row will be occupied by the number 14 for the first time?

Problem 13. When multiplying two numbers, Amy miswrote one of the factors: instead of 24 
she wrote 42, and got a product of 714. What should be the correct product?

Problem 14. In a group of 60 people, 35 have brown hair, 30 have brown eyes and 20 have both 
brown hair and brown eys. How many have neither brown hair nor brown eyes?

Problem 15. How many of the products of the numerical sequence

are divisible by 6?

Problem 16. A square is divided into 9 squares. The square R is coloured in red. Each of the 
remaining squares is coloured either in red (R), blue (B) or green (G). If in each row and in each
column the squares are coloured in all three colors, what would the colour of the X square be?

Problem 17. In a certain year, January had exactly four Tuesdays and four Saturdays. On what 
day did January 1 fall that year?

Problem 18. 111 111 111 divided by 9 equals ̅̅̅̅̅̅̅̅̅̅̅̅̅̅ Find the

Problem 19. If find .

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 C) 5 12 36 We compare

2 А) 301 The first number is , and the last is
The number is .

3 D) 12 200 – 120 = см; см < 12 dm

4 C) 45 can be read in 135 minutes, which is equivalent to 2 hours and 
15 minutes

5 A) 10,000

⏟

+ + +… + + = ⏟

Another way:

1+(2 +99)+3+(4+99 +…+ + +99) =5050 +
50 99=5050+4950=10 000.

6 D) 40 After moving the coins, in the first box there are 30 coins, and in 
the second box there are 60 coins. Before that there were 40 coins
in the first box and 50 in the second box.

7 D) 1004 If A<9, then ̅̅̅̅̅̅ ̅̅̅̅̅

In this case А= . If B<8, then ̅̅̅̅̅̅ ̅̅̅̅̅<1000
In this case B=9. ̅̅̅̅̅̅ ̅̅̅̅̅
The possible values of C are the digits 0, 1, 2,3, 4, 5, 6 and 7.

+С would be a four digit number if C=3, 4, 5, 6 and 7.
Only for C=5, 6 and 7, we get the four digit number ̅̅̅̅̅̅̅̅.
The greatest of them is , which we get from С= .

8 A) 9 The smallest number consists of the smallest amount of digits, 
therefore the predominant digits are 9.

From 2015 9= 223 (remainder 8), it follows that the smallest
number is ⏟

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9 С) 15 The minutes in 6 hours are 6 .
The hours in a day are 24.

360 24 = 15

10 B) 20 12 squares with a side of 1; 6 squares with a side of 2; 2 squares 
with a side of 3;

11

396

⏟

=12
12 19 5, ⏟

13 408 714 42 = 17; 17 24 = 408

14 15 With brown hair, but not with brown eyes: 35 – 20 = 15;
With brown eyes, but not with brown hair: 30 – 20 = 10.
With brown eyes and with brown hair: 20.

We need to deduct 15 + 10 + 20 = 45 from the total number, in
order to find the number of people that have neither brown hair
nor brown eyes: 60 – 45 = 15.

15 98 All products are divisible by 6. It is known that the product of two 
consecutive numbers is divisible by 1 2; of three consecutive
numbers – by , of four consecutive numbers –

by etc.
16 Red

or green

R B G R G B

B G R B R G

G R B G B R

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Wednesday, Thursday and Friday. January 1 is on Wednesday.

Other way

If January 1 is:

On Monday:

The Mondays would be – 5; Tuesdays – 5, Wednesdays – 5;
Thursdays – 4, Fridays – 4; Saturdays – 4, Sundays – 4;
On Tuesday:

The Mondays would be – 4; Tuesdays – 5, Wednesdays – 5;
Thursdays – 5, Fridays – 4; Saturdays – 4, Sundays – 4;
On Wednesday:

The Mondays would be – 4; Tuesdays– 4, Wednesdays – 5;
Thursdays– 5, Fridays– 5; Saturdays – 4, Sundays – 4;
On Thursday:

The Mondays would be – 4; Tuesdays – 4, Wednesdays – 4;
Thursdays – 5, Fridays– 5; Saturdays – 5, Sundays – 4;
On Friday:

The Mondays would be – 4; Tuesdays – 4, Wednesdays – 4;
Thursdays – 4, Fridays – 5; Saturdays – 5, Sundays – 5;
On Saturday:

The Mondays would be – 5; Tuesdays – 4, Wednesdays – 4;
Thursdays – 4; Fridays – 4; Saturdays– 5, Sundays– 5;
On Sunday:

The Mondays would be – 5; Tuesdays – 5, Wednesdays – 4;
Thursdays – 4; Fridays – 4; Saturdays – 4, Sundays – 5;.

18 9 12345679 9=111 111 111

19 1

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WINTER 2016: GROUP 5 
Problem 1.

A) 0 B) 4 C) 8 D) other

Problem 2. Which of the following products is the greatest?

A) B) C) D)

Problem 3. The sum of five different odd natural numbers is 27. Which of these numbers is the 
greatest?

A) 5 B) 7 C) 9 D) 11

Problem 4. How many even numbers are there from 205 to 2,017?

A) 1,812 B) 1,813 C) 907 D) 906

Problem 5. If we diminish the dividend 10 times and increase the divisor 10 times, what would 
happen to the quotient?

A) it will be diminished 100 times B) it will not change 
C) it will be diminished 10 times D) it will increase 10 times

Problem 6. By how many are the non-coloured squares less than the coloured squares?

A) 8 B) 6 C) 4 D) 2

Problem 7. Find the difference of the smallest number that is greater than 2,016 and has the 
same sum of its digits as 2,016, and the number 2,016.

A) 9 B) 7 C) 5 D) 3

Problem 8. Steve had a bowl with some sweets in it. At first he ate a third of the sweets. After 
that he ate a fourth of what was left in the bowl. In the end, he ate a sixth of the remaining
sweets. At this point there were 10 sweets left in the bowl. How many sweets were there in the
beginning?

A) 27 B) 24 C) 21 D) 18

Problem 9. What is the sum of the missing digits in the following equation?

A) 29 B) 27 C) 24 D) 18

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A) 2 B) 5 C) 10 D) 15

Problem 11. The natural number A would be increased 11 times if, on its right side, we write 
down one of the following nine digits: 1, 2, 3, 4, 5, 6, 7, 8 or 9. How many digits does the
number A have?

Problem 12. In how many different ways can we divide a set of 7 different weights (from 1 to 7 
grams each) in two groups of equal weight?

Problem 13. Place the digits 0, 1, 2 and 3 in the squares in such a way that would result in the 
greatest possible product. What is the product?

 

Problem 14. When x is divided by 55 the remainder is 22. What is the remainder when is
divided by 55?

Problem 15. A rectangle has a width of 18 cm, and a length four times greater than the width. 
How many dm is the parameter of the rectangle?

Problem 16. – = ?

Problem 17. One of the three brothers A, B and C took the golden apple. Their father asked them 
who took it and they answered as follows:

A: “B took the golden apple.” 
B: “I took the golden apple.”
C: “A took the golden apple.”

Who actually took the golden apple, if only one of the three brothers was telling the truth?

Problem 18. How many odd natural numbers that are smaller than 15 can be presented as a sum 
of two prime numbers?

(Hint: A prime number is a number, larger than 1, that can only be divided evenly by itself and 1. 
For example: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ...)

Problem 19. There were 18 apples in one basket, and 20 apples in another basket. I took a few 
apples from the first basket, and then I took as many apples as were left in the first basket, from
the second basket. How many apples in total are left in both baskets?

Problem 20. With a single jump, a grasshopper can move by 8 cm and 1 mm, and with

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 B First way: 892 440 448=892 888 = 4.

Second way: (223 110 112) 4 = 1 4 = 4.
2 C 123 5 7 > 123 5 6

123 5 7 < 123 6 7

123 6 7 = 123 42 > 123 40 = 123 5 8
The greatest product is

3 D 1 + 3 + 5 + 7 + 9 = 25, therefore the numbers are 1, 3, 5, 7 and 11. The greatest
of these numbers is 11.

4 D First way:

Let us observe the pairs of numbers – odd, even (205, 206); (207, 208);…;
(2015, 2016); (2017, 2018).

The pairs are 907. In each of them there is an even and an odd number. The
odd numbers are 907, and the even numbers from 205 to 2017 are 906.

Second way:

The even numbers from 205 to 2017 can be presented as follows:

206 + 0 2; 206 + 1 2; 206 + 2 2; ...; 206 + 905 2 = 2016 
5 A (a 10) (b 10) = (a b) 100

6 D The coloured squares are 16.

The non-coloured squares are 14: 9 squares with a size of 1 1, 4 squares with
a size of 2 2, 1 square with a size of 3 3.

7 A The smallest number we are looking for is 2,025. The difference we are
looking for is 9.

8 B We can reach the solution by carrying out a check by using the possible
answers.

If the sweets are 24, then Steve would have eaten 8 sweets, and there would
have been 16 sweets left.

After that he would have eaten 4, and 12 sweets would have been left. In the
end he would have eaten 2, and 10 sweets would have been left.

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10 B The digit b is equal to 5. The digit a is equal to 3, 4, 7, 8 or 9.

11 1 If we present the number as ̅̅̅̅ x being one of the digits from 1 to 9, then 
10A + х = 11A. Therefore x=A, i.e. A is a one-digit number.

12 4 The total weight of the set of weights is 28 grams. If the weight of 7 grams is
in the first group, then we would need to add other weights in the group that
have a total weight of 7 grams. 7 = 6 + 1 = 2 + 5 = 3 + 4 = 1 + 2 + 4, therefore
we could group them in 4 ways.

First way: first group – 7, 6, 1; second group – 2, 3, 4, 5 
Second way: first group – 7, 5, 2; second group – 1, 3, 4, 6
Third way: first group – 7, 4, 3; second group – 1, 2, 5, 6
Fourth way: first group – 7, 4, 2, 1; second group – 3, 5, 6.
13 630 30 21 = 630

14 11 The number is 55A + 22.

Therefore is 165A + 66 = 55 (3A+1) + 11.

15 18 The length is 72 cm. The parameter is 180 cm = 18 dm. 
16 252 (999 + 111 102) 4 = 252.

17 А A and B claim the same thing. From the condition of the problem it follows
that they are not telling the truth. Only C is telling the truth.

18 4 5 = 2 + 3; 7 = 5 + 2; 9 = 7 + 2; 13 = 11 + 2

19 20 First way: If we take 2 apples, then the apples left in the first basket would be
16. Then if we take 16 apples from the second basket, 4 apples will be left
there. The total number of apples left is 20.

Second way: If we take x number of apples from the first basket, then the 
apples left in it would be 18 x. Then if we take 18 x from the second 
basket, there would be 20 (18 x) = 2 + x apples left in it.

The total number of apples left is 18 – x + 2 + x = 20.

20 81 The grasshopper will go a distance of 8 81 mm, and the turtle would go a 
distance of x 8 mm.

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SPRING 2016: GROUP 5

Problem 1. The tens and hundredths digits in the number A were reversed and the resulting 
number turned out to be 20.16. What was the original number?

A) 60.12 B) 61.02 C) 20.61 D) 10.26

Problem 2. Instead of being reduced 10 times, a number was increased 10 times and the result 
was 20.16. What is the number that was meant to be received?

A) 201.6 B) 2.016 C) 0.2016 D) other

Problem 3. A is the smallest natural number, so that when divided by 9 leaves a remainder of 6. 
What would be the remainder if the number A is divided by 4?

A) 0 B) 1 C) 2 D) 3

Problem 4. The speed of a boat going along the stream is 18 km/h, and the speed of the same 
boat going against the stream is 12 km/h. What is the speed of the boat in still water?

A) 13 km/h B) 14 km/h C) 15 km/h D) 30 km/h

Problem 5. Calculate the value of the expression

.

A) 0.375 B) 0.275 C) 0.125 D) 0.1

Problem 6. The number of positive integers from 1 to 2,016, which cannot be divided by 2 or 5, 
is:

A) 1210 B) 1008 C) 202 D) 806

Problem 7. How many proper irreducible fractions are there, which have a one-digit 
denominator and a numerator other than 0?

A) 25 B) 27 C) 30 D) 35

Problem 8. If the square is magical, find .

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Problem 9. The number ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ consists of 12 digits (1, 2, 3, ..., 8 and the number is
included 4 times), and it is divisible by 3 and 5. What is the digit ?

A) 0 B) 3 C) 5 D) 9

Problem 10. The rectangle ABCD below is made up of five identical rectangles. How many 
square centimeters is the area of the rectangle ABCD equal to, if BC = 1.5 cm?

A) 3.75 B) 4.75 C) 3.5 D) 3

Problem 11. If 1 + 12 + 123 + 1,234 +...+ 12,345,678 + 123,456,789 = ̅̅̅̅̅̅̅̅̅ , then ̅̅̅̅̅ ...
Problem 12. We have 5 identical chocolate bars, each consisting of 28 pieces. We have to divide 
them equally between 7 children. What is the minimum number of times we need to break each
chocolate bar in order to do this?

Problem 13. The natural number A has 3 divisors (natural numbers, including 1 and the number

A itself), the natural number B has 2 divisors (natural numbers, including 1 and the number B 
itself), and the smallest common denominator of the two numbers is 9. How many natural
numbers are there which are divisors of the number equal to A + B (including 1 and the number 
itself)?

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Problem 16. Four children met together: Adam, Bobby, Charley and Daniel. Adam shook hands 
with 3 of these children, Bobby shook hands with 2, and Charley shook hands with 1. How many
of the children’s hands did David shake?

Problem 17. In a sports club there are 12 gold, 14 silver and 13 bronze medalists. There are 30 
individual medalists in total in the club and each of them has at least one medal. None of the gold
medalists has a silver medal, but 5 of them have bronze medals too. How many of the bronze
medalists also have silver medals?

Problem 18. Annie has a magical necklace. Each bead of the necklace is numbered (1, 2, 3, 4 
and so on). If between the beads numbered as 5 and 15 there is the same number of beads, what
is the total number of beads on Annie’s necklace?

Problem 19. The expression we are going to use for this problem is 
.

Replace one of the numbers from the expression with such a number that the initial value of the
expression would be increased by 1. How many of the numbers can be changed?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 A In the number 20.16 the digit of tens is 2, and the digit of hundredths is 6.
Therefore the number we are looking for is 60.12.

2 C The number 20.16 is 10 times greater than the given number. Therefore
the given number is 20.16 10 = 2.016. If we decrease the number 10
times we would get 2.016 10 = 0.2016.

3 C The smallest natural number which, when divided by 9 leaves a remainder
of 6, is 6. Therefore 6 4 = 1 (remainder of 2).

4 C The sum of 18 and 12 is equal to the doubled speed of the boat in still
water.

Therefore the speed of the boat in still water is equal to 30 = 15 km/h.

5 A

6 В The natural numbers from 1 to 2016 are 2016.

1008 of them are even, and 202 end in the 5 - numbers 1.5, 3.5, 5.5, 7.5,
401.5, 403.5.

The numbers which are divisible either by 2, or by 5, are 1210 in total.
The numbers which are NOT divisible neither by 2, nor by 5, are 806.

7 В The number of all proper fractions with one-digit denominators is 1 + 2 +
3 + 4 + 5 + 6 + 7 + 8 = 36.

The reducible fractions among them are: .
The number of irreducible fractions is 36 – 9 = 27 in total.

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Y X

⇒

9 A The number would be divisible by 5 if a = 0 or = 5. 
The number would be divisible by 3 if a = 0

That is only possible if = 0.

10 A The sides of each of the five identical rectangles are 1.5 cm and

1.5 . ThenAB = 0.5 + 1.5 + 0.5 = 2.5 cm andthe area of the
rectangle ABCD = 1.5

11 205 The last digit is 5, because the sum of 1, 2, 3, 4,...8 and 9 is 45.

The digit before last would be the last digit of the sum 1+2+3+...+7+8 plus
4, i.e. 36 + 4 = 40. The digit before last is 0.

The other digit we are looking for is equal to the last digit of 1+2+3+...+7
plus 4, i.e. 28 + 4 = 32. The last three digits are 2, 0 and 5. ̅̅̅̅̅ 205

12 6 The number of all pieces of all five chocolates is 5 28=140.
Therefore each child should get 140 7 = 20 pieces.

By breaking one chocolate, we can get 20 pieces and have 8 left.

In this way we can give 20 pieces to 5 children, however there would be 2
more children and 5 more parts, each consisting of 8 pieces, left.

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13 6 The number B is a simple number and is a divisor of 9. Therefore B = 3 
and A = 9, A + B = 12, and the natural numbers which are divisors of 12 
are 6: 1, 2, 3, 4, 6 and 12.

14 5 The numbers smaller than 2016 are 6: 1026, 1062, 1206, 1260, 1602,
1620;

The numbers greater than 2016 are 11: 2061, 2106, 2160, 2601, 2610,
6012, 6021,6102, 6120, 6201, 6210.

15 3 The possible options are:
4 and 5, product 20;
6 and 7, product 42;

8 and 9, product 72.

16 2 A B C D

A + + +

B + – +

C + – –

D + + –

If we add the number of hand shakes, the number must be divisible by 2,
because each hand shake is counted twice.

In this case the number of hand shakes is 6 + x.

We can mark the number of David’s handshakes with x. The number x can 
NOT be greater than 3.

6 + x can be divided by 2 only if x is either 0 or 2.

However, x is not 0, because Adam shook hands with all the children. 
Therefore x = 2. David shook hands with 2 children.

17 4 Let us subtract 12 gold medalists from the total number of medal winners.
Therefore 18 people have won bronze and silver medals.

The bronze medalists who have no gold medals are 13 – 5 = 8.
The silver medalists are 14. We get 14 + 8 = 22 in total.

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18 20 The beads with numbers from 6 to 14 are situated between the beads with
numbers from 5 to 15. The beads are 9 in total. The beads on the opposite
side are also 9. If we also note the 2 beads numbered 5 and 15 we will find
that the beads on Annie’s necklace are 2 9 + 2 = 20.

19 2 In order for the value of the expression to be increased by 1, the following
needs to be true:

The first number in the expression 3 + 12 – 10 needs to be exchanged with
4. Then the initial value would be increased by 1.

 2 = 4 would be possible if we exchange 6 for 8. 
6  = 4 is not possible.

The second number in the expression 3 + 12 – 10 needs to be exchanged
with 13. Then the initial value of the expression would be increased by 1.
 3 = 13 is not possible.

4  = 13 is also not possible.

The third number in the expression 3+12-10 needs to be exchanged with 9.
Then the initial value would be increased by 1.

1  = 9, if we exchange 10 for 9.
 10 = 9 is not possible.

We can exchange two of the numbers in the expression, so that the initial
value would be increased by 1:

8 2 + 4 3 – 1 10
6 2 + 4 3 – 1 9

20 2 We would have to place 3 coins on each pan of the scales.

If the scales are balanced, then we would have to place 1 of the 
remaining coins on each pan of the scales and if the scales are still
balanced, the third remaining coin is fake; if it is not balanced, the lighter
(fake) coin can be found on the higher pan of the scales.

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FINAL 2016: GROUP 5

Problem 1. Find the sum of the fractions and , if  .

A) B) C) D) or

Problem 2. The numbers from 0 to 40 are written down one after another: 
01234567891011...37383940.

In how many ways can we pick out two consecutive digits, so that their sum would be 10?

A) 4 B) 5 C) 6 D) 7

Problem 3. There are several points along a straight line. A student placed a point between every 
two adjacent points. After doing this a number of times, there were now 129 points along the
straight line. How many points were possibly on the straight line originally (before the student
placed any extra points)?

A) 2 B) 3 C) 4 D) 6

Problem 4. Adam has 44 marbles - blue, red, white and yellow. The number of the blue marbles 
is 2 more than that of the red, the number of the red marbles is 4 more than that of the white, and
the number of the white marbles is 6 more than that of the yellow. How many blue marbles does
Adam have?

A) 12 B) 14 C) 16 D) 18

Problem 5. Iva arrived at the bus stop and looked at her watch, which showed 08:01h. It meant 
that she was 2 minutes late for her bus. What she did not know was that her watch was running 5
minutes ahead. If the bus came 1 minute late, for how many minutes would Iva have to wait at
the bus stop?

A) 4 B) 5 C) 3 D) 6

Problem 6. How many four-digit numbers are there that can be written down using the four 
digits 1, 2, 3 and 4, in such a way that 1 is not the digit of ones, 2 is not the digit of tens, 3 is not
the digit of hundreds, and 4 is not the digit of thousands?

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Problem 7. What is the number in which the digit of tenths is smaller than the digit of tens?

A) 222.31 B) 209.09 C) 32.32 D) 345.255

Problem 8. A storage room can be filled up with either 12 chests, or with 18 boxes. There are 
currently 4 chests and 9 boxes in the room. How many more boxes can fit into the room?

A) 6 B) 4 C) 2 D) 3

Problem 9. A spring with a flow rate of 84 liters of water per minute provides water for three

fountains. Four times more water reaches the second fountain than does the first one, and half as
much water reaches the third fountain than does the second one. How many liters per minute is 
the flow rate of the fountain which receives the least amount of water?

A) 4 B) 7 C) 12 D) 14

Problem 10. Which of the following fractions is equal to ?
A)

B)

C)

D)

Problem 11. I added each two of the numbers A, B and C, and then added the sums again. At last 
I got . What is A + B + C equal to?

Problem 12. If the dividend is , and the divisor is

, then the quotient is a number that

is written down using X different digits. Calculate X.

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Problem 14. Ten teams are participating in a football tournament. Each team plays every other 
team exactly once. For each match, the winning team gets 3 points, the losing team gets 0 points,
and in case of a draw each team gets 1 point. At some point in the tournament it turns out that the
teams have earned a total of 131 points. How many games are yet to be played?

Problem 15. The two-digit numbers ̅̅̅, ̅̅̅ and ̅̅̅ are multiples of 17 (each letter represents one
number). Calculate the greatest possible value of

Problem 16. How many of the four-digit numbers written down using the four digits 2, 0, 1, 6 
are divisible by 36?

Problem 17. A cuboid has been formed from two identical cubes with a surface area of 1.5 sq.

cm each. Calculate the surface area of this cuboid.

Problem 18. The fraction is presented as an infinite repeating decimal. What are the digits
that are not used when writing it down?

Problem 19. How many prime numbers are there, smaller than 99, for which is also a
prime number smaller than 99?

Problem 20. The numbers A and B are such that 143 A + 325 B = 6.5. Calculate the value of 
the following expression if A and B are the same as above:

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 D  ⇒ or  .

2 C

0123456789101112131415161718192021222324252627282930313233343
53637383940

The digit couples are: 91, 19, 28, 82, 37, 73.

3 B

If the number of the points is 3, we will place another 2; now there are 5
points and then we will place 4 more points to get 9. Then we will place
another 8 points to get 17. Then we will place another 16 to get 33. Then
we will place another 32 to get 65. Then we will place another 64 and the
final number of the points will be 129.

4 C

If the yellow marbles are 4, then the white marbles are 10, the red are 14,
and the blue are 16. The total number is 4 + 10 + 14 + 16 = 44.

5 А

The watch showed that the time was 08:01 h. She was 2 minutes late, which
meant that she was supposed to arrive at 07:59 h, according to her watch.
This meant that she came 3 minutes early because the bus (according to her
watch) was supposed to arrive at 08:04 h. The bus was running 1 minute
late. Therefore it would arrive at 08:05 h. Iva would have to wait for 4

minutes.

6 B

We can choose either 1, 2 or 3 for the digit of thousands. 1, 2 or 4 for the
digit of hundreds; 1, 3 or 4 for the digit of tens and 2, 3 or 4 for the digit of
ones.

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If the digit of thousands is 3, then the numbers are 3412, 3214 and 3142.
The number of four-digit numbers are 9 in total.

7 D

In the number 345.255, the digit of tenths is 2 and is smaller than the digit
of tens, which is 4.

8 D

of the room is occupied. The unoccupied part is equal to ,

which means there is still space for 3 boxes.

9 C

We can carry out a check using the possible answers. If the flow rate of the
second fountain is 48 liters per minute, then the flow rates of the first and 
third fountains would respectively be 12 liters per minute and 24 liters per 
minute. In this case 48 + 12 + 24 = 84.

10 A

The greatest common divisor of 4095 and 6426 is 63. After cancelling the
fraction, we would get

as a result.

11 or

1.2 ⇒ .

12 8 The quotient is 12 345 679.

13 1/3

Alex is able to pay 3 (1 + 2 + 5 + 10 + 20 + 50) = 264 cents.

3.96 – 2.64 = 1.32, so his father needs to pay 132/396=1/3 of the book
price.

14 0 or 1

The total number of matches is 45 and the maximum points a team can earn
is 135. Now since 131 points have been earned, there are 4 points left.
Therefore 131 points can be earned in three ways:

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15 21

The number ̅̅̅ can be 17, 34, 51, 68 or 85. The number ̅̅̅ can be 17, 34,
51, 68 or 85. We must keep in mind that the second digit of ̅̅̅ is also the
first digit of ̅̅̅ Therefore the options are either or .

1. ⇒ ⇒
2. ⇒ ⇒

The sum we are looking for is 21.

16 6

The sum of the digits is 9, so all of the four-digit numbers written down
using these digits are divisible by 9. In order for it to be divisible by 36, it
also needs to be divisible by 4. In this case the last two digits must be
arranged as follows: хх6 , хх2 , хх 6 and хх 2. It is now possible to find
the four-digit numbers we are looking for: 1260, 2160, 6120, 1620, 2016
and 6012.

17 2.5

The length of the edge of the cube is 0.5 cm. We can find the surface area of 
the cuboid by subtracting the doubled area of one of the sides of the cube
from the sum of the surface areas of the two cubes: 2 1.5 – 2 0.5 0.5
= 3 – 0.5 = 2.5 sq. cm.

18 3, 6 and
9

̅̅̅̅̅̅̅̅̅̅

The digits 3, 6 and 9 are not used to write down the repeating decimal.

19 7 (2; 5), (3; 7), (5; 11), (11; 23), (23; 47), (29; 59), (41; 83).

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TEAM COMPETITION – NESSEBAR, BULGARIA

MATHEMATICAL RELAY RACE

The answers to each problem are hidden behind the symbols @, #, &, § and * and are used in 
solving the following problem. Each team, consisting of three students of the same age group,
must solve the problems in 45 minutes and then fill a common answer sheet.

GROUP 5

Problem 1. Subtract the number @ from the numerator and from the denominator of the fraction

, in order to get a fraction equal to . Calculate @.

Problem 2. We have made a random selection of # three-digit numbers. Among all three-digit 
numbers, there would always be at least 3 which are co-prime with @. Find the smallest possible 
value of #.

Problem 3. The children from a school class had to solve # problems for homework. Three of 
them solved respectively 60, 50 and 40 problems. At least & problems have been solved by all 
three of them. Find &.

Problem 4. Two ants started walking towards each other simultaneously from the two points A 
and B. One of the ants will travel the distance in & hours and the other one will travel the 
distance two hours faster. What part of the full distance would they need to walk before they

meet, if two hours have passed since their departure? Denote the answer using §. Find §.

Problem 5. If §

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 @ = 3

2 # = 72

The numbers which are not coprimes to 3 are 69. Those numbers are: 
3 8 = 4, 3 9 = 7, 3 = 3 ,..., 3 76 = 988.

The number we are looking for in this case is 69 + 3 = 72.

3 & = 6

The first student did not solve 2 problems, the second did not solve
22, and the third did not solve 32 problems. If none of the other two
students solved the unsolved problems, then the unsolved problems
would be 66 in the worst case scenario. In this case 72 – 66 = 6
problems have been solved by the three of them.

4 § = They have walked

of the full distance.

The remaining distance they have yet to walk is

5 * =

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MATHEMATICS WITHOUT BORDERS 
2015-2016

AUTUMN 2015: GROUP 6

Problem 1. 1,000,000 – 100.1 =

A) 999,998.9 B) 999,989.99 C) 900,000.9 D) 1,000,010.01 
Problem 2. Find the value of the expression

A) 5 B) 1 C) 0,9 D) other

Problem 3. When dividing the sum of 3 consecutive odd numbers by 6, the remainder we get is 
always:

A) 0 B) 1 C) 2 D) 3

Problem 4. If the three-digit numbers ̅̅̅̅̅̅ and ̅̅̅̅̅ is divisible by 9, what is the number Y?

А) 2 B) 3 C) 4 D) 5

Problem 5. Peter reads 15 pages in 20.5 minutes. How long will it take him to read 16 pages at that 
rate?

A) 21 min 33 sec B) 21 min 52 sec C) 20.33 min D) 21.52 min

Задача 6. In the table the numbers are placed so that all rows, columns and the two diagonals will 
sum to the same value. Which number you should place instead of „?“ ?

1.2

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Problem 7. Two fractions divide the interval with the boundaries of and into three equal parts.
The smaller of the two fractions is:

A) B) . C) D)

Задача 8. The price of a product has been changed twice, either increased or decreased. In which 
case we can buy the product at the lowest price?

A) If first decreased by 10%, then increased by 10%

B) If first increased by 15 %, then decreased by 15 % 
C) If first increased by 20 %, then decreased by 20 % 
D) If first decreased by 25 %, then increased by 25%

Problem 9. Find the tens digit of the value of the expression

A) 0 B) 1 C) 2 D) 5

Problem 10. In a group of 80 people, 39 have brown hair, 30 have brown eyes and 15 have both 
brown hair and brown eys. How many have neither brown hair nor brown eyes?

A) 15. B) 26 C) 31 D) other

Задача 11. All four-digit numbers divisible by are formed by the digits 0, 2, 3 and 7.
Determine the number of possible values of x + y + z, if x, y and z are positive integers.

(Hint: ⏟ ⏟ ⏟ )

Problem 12. There were a total of 90 coins in two boxes. A third of the coins from the first box 
were then shifted into the second one. As a result, the number of coins in the second box was twice
as much as the number of coins in the first one. What was the number of coins in the first box
before the shift?

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Problem 14. Find the value of the expression

Problem 15. A rectangular sheet of size 3 cm by 4 cm is cut into squares with side lengths that are 
whole numbers. If the sheet is cut into the smallest possible number of squares, how many squares
of side length 1 cm are there?

Problem 16. How many of the products of the numerical sequence

are divisible by 24?

Problem 17. How many are the 4-digit numbers which end in 3 and are divisible by 3?

Problem 18. In a certain year, February had exactly five Saturdays. On what day did March 1 fall 
that year?

Задача 19. If then the value of is….

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 B

2 B

3 D We can easily find the answer by observing an example:

1 + 3 + 5 = 9; 9 6 = 1(remainder of 3).

If the first odd number is 2n+1, the next two would be 2n+3 and 2n+5. 
Their sum is 6n+9. The quotient is 2n+1, remainder – 3.

4 D The sum of the digits of ̅̅̅̅̅ is a number that is a multiple of 9, if X = 3.
The sum of the digits of the number ̅̅̅̅̅̅ for Х = 3 is 4 + Y is a number
that is a multiple of 9, if Y = 5.

5 B 15 pages in 20,5 minutes = 1230 seconds. One page can be read in 82
seconds = 1 minute and 22 seconds.

16 pages = 15 pages +1 page

16 pages can be read in 20 minutes and 30 seconds + 1 minute and 22
seconds = 21 minutes 52 seconds.

6 C Let x denote the number n the second row, third column. We reach the 
conclusion that: .

7 A From ; it follows that the interval is If we divide it into three 
parts, we will find that each part is

The smaller of the two fractions is

8 D The price of the goods is respectively 0,99; 0,9775; 0,96; 0,9375 of the
initial price.

9 A

The number is a multiple of 100, therefore the digit of tens is 0.
10 B With brown hair, but not with brown eyes: 39 – 15 = 24;

With brown eyes, but not with brown hair: 30 – 15 = 15.
With brown eyes and with brown hair: 15.

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number of people who neither have brown hair nor brown eyes:
80 – 54 = 26.

11 2 The sum of the digits of each of the numbers we are looking at is 12. The
number is divisible by 3, but not by 9. Therefore y = 1.

number x, y, z x+ y+ z

2370 3 790=3 5 194=2 3 5 97
x=1 ; y=1; z=1

2730 3 910 =3 5 182=2 3 5 91
x=1 ; y=1; z=1

3270 3270=3 5 218=2 3 5 109
x=1 ; y=1; z=1

3720 3720=2 3 5 124=8 3 5 31
x=3 ; y=1; z=1

7230 7230=2 3 5 241=2 3 5 241
x=1 ; y=1; z=1

7320 7320=2 3 5 244=8 3 5 61
x=3 ; y=1; z=1

12 45 It is a good idea to solve the problem by starting at the end.

The coins at the end are respectively 30 and 60, in the first and second
boxes. 2/3 of the initial coins are left in the first box. Therefore the coins
in the first box were 45.

13 5 Solution. The position of the points is ABC, ACB, BCA or CBA. The first
position is not possible, because AB=16cm>AC=8cm. CBA is not possible
either because AB=16 cm>AC=8 cm. If the position is ACB, then the
distance between the midpoints of BC and AB is 5cm.

If the position is BCA, then the distance between the midpoints of BC and
AB is 5cm.

14 36

⏟

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squares with a side of 1cm.

16 97 The product of 4 consecutive numbers is divisible by
17 300 The 4 digit numbers that end with 3 would be divisible by 3 if the 3 digit

number that we got after removing the digit of ones of the 4 digit number
is divisible by 3. The total number of 3 digit numbers is 999 – 99 = 900.
Each third number is divisible by 3.

Their number is 300. 
18 Sunday If 1st of February is on a:

Monday:

The Mondays would be – 4 or 5; Tuesdays – 4, Wednesdays – 4;
Thursdays – 4, Fridays – 4; Saturdays – 4, Sundays – 4;

Tuesday:

The Mondays would be – 4; Tuesdays – 4 or 5, Wednesdays – 4;
Thursdays – 4, Fridays – 4; Saturdays– 4, Sundays – 4;

Wednesday:

The Mondays would be – 4; Tuesdays –, Wednesdays – 4 or 5; Thursdays

– 4, Fridays– 4; Saturdays – 4, Sundays– 4;

Thursday:

The Mondays would be – 4; Tuesdays – 4, Wednesdays – 4 ; Thursdays –
4 or 5, Fridays – 4; Saturdays – 4, Sundays – 4;

Friday:

The Mondays would be – 4; Tuesdays – 4, Wednesdays – 4; Thursdays –
4, Fridays – 4 or 5 ; Saturdays – 4, Sundays – 4;

Saturday:

The Mondays would be – 4; Tuesdays – 4, Wednesdays – 4; Thursdays– 4,
Fridays– 4; Saturdays – 4 or 5, Sundays – 4;

Sunday:

The Mondays would be – 4; Tuesdays – 4, Wednesdays – 4; Thursdays –
4, Fridays – 4; Saturdays – 4, Sundays – 4 or 5;

19

( )

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WINTER 2016: GROUP 6 
Problem 1. Which of the following equalities is correct?

A) B) C) D)

Problem 2 What is the value of the expression ?

A) 1 B) C) D) 2

A) 36 B) C) 4 D)

Problem 4. How many even numbers are there with an absolute value smaller than 10?

A) 4 B) 8 C) 9 D) other

Problem 5. What is the remainder left when is divided by 15?

A) 0 B) 5 C) 10 D) 15

Problem 6. The product of two integers smaller than 7 and greater than ( 77) is 77. The sum of
these numbers is:

A) B) C) 18 D) 78

Problem 7. When apples are dried they lose 84% of their weight. How many kilos of apples would 
be needed to produce 24 kg of dried apples?

A) 84 B) 100 C) 125 D) 150

Problem 8. Steve had a bowl with some sweets in it. At first he ate a third of the sweets. After that 
he ate a fourth of what was left in the bowl. In the end, he ate a sixth of the remaining sweets. The
initial number of sweets CANNOT be:

A) 12 B) 24 C) 30 D) 36

Problem 9. The code for a safe is made up of all digits that are multiples of 3, without any digit 
being repeated.What is the maximum number of failed attempts that we would need to make before
breaking the code?

A) 5 B) 6 C) 23 D) 24

Problem 10. Two identical rectangles with a length of 4 cm and a width of 3 cm overlap, forming a 
square. How many sq. cm is the area of their common part (the part where they overlap)?

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Problem 11. The natural number A would be increased 11 times if we write down on the right side 
of it one of the following nine digits: 1, 2, 3, 4, 5, 6, 7, 8 or 9. How many digits does the number A 
have?

Problem 12. In how many different ways can we divide a set of 7 different weights (from 1 to 7 
grams each) in two groups of equal weight?

Problem 13. The natural numbers from 1 to N have been recorded next to one another. The result is 
a multi-digit number containing 3 N digits. What is the number N?

Problem 14. What is the smallest natural number N, for which the product of 13, 17 and N can be 
presented as the product of three consecutive natural numbers?

Problem 15. Calculate the following:

Problem 16. Three fishermen go fishing regularly. The first one goes every day, the second one

goes once every three days, and the third one goes once every four days. Today is Sunday and
they’re all at the lake, fishing. In how many days, counting from Monday onwards, are they all
going to be together at the lake again?

Problem 17. One of the three brothers A, B and C took the golden apple. Their father asked them 
who took it and they answered as follows:

A: “B took the golden apple.” 
B: “I took the golden apple.”
C: “A took the golden apple.”

Who actually took the golden apple, if only one of the three brothers was telling the truth?

Problem 18. What is the 2016th digit after the decimal point of the infinite periodic decimal equal 
to the fraction

Problem 19. There were 18 apples in one basket, and 20 apples in another basket. I took a few 
apples from the first basket, and then I took as many apples as were left in the first basket, from the
second basket. How many apples in total are left in both baskets?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 C 2 + 3 2 = 4 ≠ 2

(the only correct equality)

(division by 0 is not possible)

2 B

3 D 1 2 + 3 – 4 + 5 – 6 + 7 8 4

4 C The numbers are 8, 6, 4, 2, 0, 2, 4, 6 и 8.

5 C quotient + remainder

Hence the remainder is a natural number that is a multiple of 5, but is smaller
than 15, i.e. it is either 0, 5 or 10.

The remainder is not 0, because is not divisible by 3.

The remainder is not 5, because in this case
The number equal to = 1 ⏟

is not divisible by 3.

Therefore the remainder is 10.

6 А 77 = ( 1) ( 77) = ( 7) ( 11) = 1

Among these, two integers are smaller than 7 and greater than 77, hence the
numbers we are looking for are 7 and 11. Their sum is 18.

7 D 100 % 84 % = 16 %
16% of x is 24 kg

We need 150 kg of apples to produce 24 kg dried apples.

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by 12 (the least common multiple of 3, 4 and 6).
The only number that is not divisible by 12 is 30.

9 C The multiples of 3 are the digits 0, 3, 6 and 9. Therefore all possible codes
would be 4 3 2 1 = 24.

In the worst case scenario, we would break the code after making 23 failed
attempts. The code would be the 24th possible number.

10 B The square must have a side of 4 cm. Then the common part would be a 
rectangle with a width of 2 cm and a length of 4 cm.

The area would then be 8 sq. cm

11 1 If we write down the number ̅̅̅̅ where x is one of the digits from 1 to 9, then 
10A + x = 11A. Then x = A, i.e. A is a one-digit number.

12 4 The total weight of the set of weights is 28 grams. If the weight of 7 grams is

in the first group, then we would need to add other weights in the group that
have a total weight of 7 grams. 7 = 6 + 1 = 2 + 5 = 3 + 4 = 1 + 2 + 4, therefore
we could group them in 4 ways.

First way: first group – 7, 6, 1; second group – 2, 3, 4, 5

Second way: first group – 7, 5, 2; second group – 1, 3, 4, 6

Third way: first group – 7, 4, 3; second group – 1, 2, 5, 6

Fourth way: first group – 7, 4, 2, 1; second group – 3, 5, 6.

13 1,107 9 1 + 90 2 + 900 3 + (N 999) 4 = 3N

N = 1,107

14 600 13 4 17 3 = 52 51, therefore the smallest natural number we are
looking for is 4 3 50 = 600.

13 17 600 = 50 51 5

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16 11 Let us assume that today is Sunday and all three fishermen are fishing together.
The first fisherman goes fishing every day.

The second fisherman will go fishing on Wednesday, Saturday, Tuesday,
Friday.

The third fisherman will go fishing on Thursday, Monday, Friday.
There are 11 days from Monday to Friday the week after.

They would all be fishing together again in 11 days.

17 А The claims of A and B are identical. From the condition of the problem it 
follows that they are not telling the truth. Only C is telling the truth: “A took 
the golden apple.”

18 3

= 0.232323...

As 2016 is an even number, the 2016th digit after the decimal point is 3.

The total number of apples left is 18 x + 2 + x = 20.

20 15 105 = 3 × 7 × 5 = 1 × 3 × 5 × 7

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SPRING 2016: GROUP 6

Problem 1. The value of expression is:

A) - 12 084 B) 2 014 C) 2 028 D) -2 028

Problem 2. The numerator of the fraction was increased by 20% and the denominator was
decreased by 40%. The result was a new fraction - . The quotient of the two fractions, , is
equal to:

A) 0.5 B) 0.75 C) 1.2 D) 2

Problem 3. Thus reads the postulate of Bertrand: "If n is a positive integer that is greater than 1, 
then there is always a prime number , where ". How many prime numbers are there if
= 25?

A) 5 B) 6 C) 7 D) 25

Problem 4. The average age of the crew members excluding the captain is 25 years, and including 
the captain - 26 years. If the captain is 30 years old, then the crew consists of:

A) 4 people B) 5 people C) 6 people D) more than 6 people

Problem 5. The product of the natural numbers from 1 to 122 is divisible by The greatest
possible value of the natural number is:

A) 10 B) 11 C) 12 D) 13

Problem 6. The points with coordinates of (0;0), X (2;0), (2;3) and (0;3) are vertices of the 
rectangle . Which one of the following points is outside of the rectangle?

A) A (0;0) B) B (1;1) C) C (2;2) D) D (3;3)

Problem 7. 26 litres of juice must be bottled in 10 bottles of either 1 litre, 3 litres or 5 litres. The 1 
litre bottles are of an even number. How many bottles of 5 litres are there?

A) 1 B) 2 C) 3 D) 4

Problem 8. If the square is magical, find .

A) 11 B) C) 12 D)

Problem 9. What result would you get after calculating the following expression?

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Problem 10. When water turns into ice, its volume increases by

If we defrost the ice, how much

would its volume decrease?

A) B) C) D)

Problem 11. If the numbers А and B are such that the expression | | | |, 
then A + B = …

Problem 12. The number ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ consists of 12 digits (1, 2, 3, ..., 8 and the number a is 
included 4 times), and it is divisible by 36. What is the digit a?

Problem 13. The natural number A has 3 natural numbers as its divisors (including 1 and the 
number itself), the natural number B has 2 natural numbers as its divisors (including 1 and the 
number itself), and the smallest common denominator of the two numbers is 12. How many natural
numbers are there, that are divisors of the number equal to A + B (including 1 and the number 
itself)?

Problem 14. Three consecutive natural numbers are the digits of hundreds, tens and ones of a 
three-digit number. By how much would this number be increased if we write down its three-digits in reverse
order?

Problem 15. Four children met together: Adam, Bobby, Charley and Daniel. Adam shook hands 
with 3 of these children, Bobby shook hands with 2, and Charley shook hands with 1. How many of
the children’s hands did David shake?

Problem 16. There are 9 coins, one of which is fake and it is lighter. What is the least possible 
number of times we would have to weigh the coins (using scales) in order to find the fake one?

Problem 17. How many digits are there in the number that is equal to the value of the following 
expression?

⏟

Problem 18. If the product of 7 integers is a positive number, how many different options are there 
for the possible number of negative integers among these multipliers?

Problem 19. Three points lie on a straight line. The length of each line segment is 2, 3 and k. What 
is the value of k?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 C 
2 A

3 B The prime numbers that are greater than 25 and smaller than 50 are: 29,
31, 37, 41, 43 and 47.

4 B First way: We can check: if the crew members, excluding the captain, are
4, then the sum of their ages would be 100. If we add the captain, then
they would be 5 and the sum of their ages would be 130, and 130 5=26.

Second way: Let us assume that the crew members, excluding the captain,
are equal to . The sum of their ages is 25x. Including the captain, the sum 
of their ages would be equal to 26 or 25 + 30. Thus we get the equation
26 = 25 + 30, the solution of which is = 4.

5 C The numbers from 1 to 122 which are divisible by 2 are 61. Among the
same numbers, the numbers which are divisible by 11 are 11: 11, 22,
...,88, 99, 110, 121 and their product is divisible by because
121 = 11 11. Then the product of the numbers from 1 to 122 can be

presented as follows: where A is not divisible by 11, therefore it 
is not divisible by 22.

The greatest possible value of the natural number is 12.

6 D The four points lie on the same straight line, and the point D is situated 
outside the rectangle.

7 В Let x, y and z denote the number of bottles containing 1, 3 and 5 liters. 
Then ⏟

.
We have these options to bottle the juice:

If z = 1 y = 6 x = 3; If z = 2 y = 4 x = 4; If z = 3 y = 2 x = 5. 
Since the 1 litre bottles are of an even number, only the second option is
possible – the number of 5 litre bottles is 2.

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The sum of the numbers along one of the diagonals ( 1, 8, Y) is equal to
the sum of the numbers along the column (X, 4, Y).

Therefore 1 + 8 = X +( 4) X=11. Such a magical square does exist:
14

20
5 2 17

9 C

10 C Let us assume that the volume of the water is equal to 1 kg. If it is frozen, 
it will increase its volume to 1 +

kg. The volume of the ice is equal to

kg. It would need to be decreased by x, until it reaches 1. Thus we can 
establish that

. Then the volume would be decreased by

11 | | | | | | and | |

12 0 The number must be divisible both by 9 and by 4.

The number would be divisible by 9 if = 0 or = 9.
The number ̅̅̅̅ would be divisible by 4 if = 0.

13 2 The number is a prime and a divisor of 12. Therefore is either 2 or 3.
If = 2, i.e. would have more than 3 natural numbers as its 
divisors. If = 3, . The number of natural numbers 
which are divisors of 7 is two 1 and 7.

14 198 ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅

15 2 A B C D

A + + +

B + – +

C + – –

D + + –

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because each hand shake is counted twice.
In this case the number of hand shakes is 6 + x.

We can mark the number of David’s handshakes with x. The number x can 
NOT be greater than 3.

6 + x can be divided by 2 only if x is either 0 or 2.

However, x is not 0, because Adam shook hands with all the children. 
Therefore x = 2. David shook hands with 2 children.

16 2 We would have to place 3 coins on each pan of the scales.
If the scales are balanced:

then we would have to place 1 of the remaining coins on each pan of
the scales and if the scales are still balanced, the third remaining
coin is fake; if it is not balanced, the lighter (fake) coin can be
found on the higher pan of the scales.

If the scales are not balanced, the lighter coin can be found on the higher 
pan of the scales. We would then have to compare the weights of two of
those three coins.

17 1 
because the number 0 is found among the multipliers.

18 4 The negative numbers among the multipliers are either 0, 2, 4 or 6. There
are 4 possibilities.

19 5 or 1 Let A, B and C denote the points on the straight line. If AB = 2, BC = 3 and

AC = k, the points can be located in 4 possible ways:

А, В, С 2 + 3 = k k = 5 
B, A, C 2 + k = 3 k = 1 
C, A, B k + 2 = 3 k = 1 
C, B, A 3 + 2 = k k = 5

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FINAL 2016: GROUP 6

Problem 1. The product of two integers which are smaller than 4 and greater than ( 8) is equal to
8. The absolute value of the difference of these numbers is ( ).

A) 0 B) 1 C) 2 D) 7

Problem 2. The numbers from 0 to 100 are written down in order: 
01234567891011...979899100.

If we pick out three consecutive digits, the first two of which have a sum of 10, the third digit
cannot be:

A) 1 B) 2 C) 3 D) 5

Problem 3. What is the remainder when dividing by 12?

A) 0 B) 2 C) 4 D) 8

Problem 4. Four points have been placed on the square grid below. Three of them have the 
coordinates of ( 5; 0), ( 1; 0) and (0; 0). Find the abscissa of the fourth point.

A) 4 B) 3 C) 2 D) 1

Problem 5. If , the value of the expression is 1. Calculate the value of the
expression if х = 2.

A) 4 B) 2 C) 1 D) 0

Problem 6. Adam has blue, red, white and yellow marbles. The number of the blue marbles is 2 
more than that of the red ones, the number of the red ones is 4 more than that of the white ones and
the number of the white ones is 6 more than that of the yellow ones. What is the least possible
number of the marbles that Adam has?

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Problem 7. A spring with a flow rate of 84 liters of water per minute provides water for three 
fountains. Four times more water reaches the second fountain than does the first, and half as much
water reaches the third fountain than does the second. By how many liters per minute is the flow 
rate of the second fountain greater than the flow rate of the third fountain?

A) 12 B) 18 C) 24 D) 30

Problem 8. Three fractions divide the interval between and into four equal parts. The greatest of the
three fractions is:

A)

B) C) D)

Problem 9. If the square on the right is a magic square, find .

X

A) 1 B) - 1 C) 2 D) -2

Problem 10. George chooses 4 numbers. The sums of every three of them are 13, 14, 15 and 6.
The sum of these numbers is:

A) 9 B) 12 C) 18 D) 42

Problem 11. How many integers А are there, so that = ̅̅̅̅ (where ̅̅̅̅ is a two-digit number)
and the two-digit number ̅̅̅̅ can be presented as the product of two different one-digit numbers?

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Problem 13. Alex has 3 of each of the following coins: 1, 2, 5, 10, 20 and 50 cents. He wants to 
buy a book that costs 3 euros and 96 cents but he does not have enough money, so he asks his father
for help. What fraction of the book price does his father need to pay? Write down the answer in the
form of an irreducible fraction.

Problem 14. Ten teams are participating in a football tournament. Each team plays every other 
team exactly once. For each match, the winning team gets 3 points, in the case of a draw both teams
get 1 point each, and the losing team gets 0 points, and in case of a draw each team gets 1 point. At
some point in the tournament it turns out that the teams have earned a total of 131 points. How
many games are yet to be played?

Problem 15. Calculate , if and

Problem 16. How many of the four-digit numbers that consist of all four digits 2, 0, 1 and 6, are 
divisible by 36?

Problem 17. A cuboid has been formed from three identical cubes, each with a surface area of 6 sq.

cm. Calculate the surface area of this cuboid.

Problem 18. Five weavers weave 10 rugs in three days. How many rugs will 3 weavers weave in 7

days?

Problem 19. For which primes , smaller than 99, is also a prime?

Problem 20. The numbers A and B are such that .

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 C 8 = ( 2) ( 4), therefore the difference we are looking for is
| |

2 А The triplets are

910, 192, 282, 829, 373, 739 , 464, 647, 555, 556, 646, 465, 737, 374,
828, 283, 919, 192 . The third digit is never 1.

3 С quotient + remainder. It is not difficult to realize that the
remainders are among the numbers 0, 2, 4, 6, 8 and 10.

It is impossible for the remainder to be 0 or 6, because 3 cannot divide
.

It is impossible for the remainder to be 2 or 10 either, because 4 can divide
, but can not divide 12 quotient + 2 or 12 quotient + 10.

There are two remaining options: 4 or 8.

The remainder is 4, because only is divisible by 3, but

is not divisible by 3.

4 В The three given points are all on the x-axis.

The fourth point has the coordinates of ( 3; 2), so the abscissa is -3.
5 D If we replace х with – 3, we would get:

.

So now the expression becomes . Its value is

6 C The number of the white marbles is 6 more than that of the yellow ones;
the number of the red marbles is 4 more than that of the white ones, i.e. 10
more than that of the yellow ones, and the number of the blue marbles is 2
more than that of the red ones, i.e. 12 more than the yellow ones.

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7 C If the flow rate of the first fountain is , then the flow rates of the second
and third respectively are 4 and 2 . Concerning we get the following:

We are looking for .

The flow rate of the second fountain is 24 liters more than the flow of the 
third.

8 C Have a look at the following fractions:

The fraction we are

looking for is

9 B If we compare the sums of the third row and the second column, we will
get the following: 13 + X = 19 + 5 X = 1

10 B Each of the numbers is part of three sums. In order to find the sum of the
numbers, we must add the sums and divide the result by 3. We get the
following:

( 6 + 13 + 14 + 15) 3 = 12.

11 2 Among the perfect squares 16, 25, 36, 49, 64 and 81, only 36 and 81 are
such that the numbers 63 and 18 can be presented as the product of two
one-digit numbers.

Therefore the number A can either be 6 or 9.

12 30 Let us paint the board using a chessboard pattern. The cut out squares
would be of the same color, and each of the rectangles would cover
both colors.

In this case it would be impossible to get 31 rectangles. However it is
possible to get 30. (Two squares would be unnecessary – the ones adjacent
to the cut out squares)

13 1/3 Alex paid 3 (1 + 2 + 5 + 10 + 20 + 50) = 264 cents.

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14 0 or 1 The total number of matches is 45 and the maximum points a team can
earn is 135. Now since 131 points have been earned, there are 4 points
left.

Therefore 131 points can be earned in three ways:
41 wins and 4 draws, with no matches to go;
43 wins and 1 draw, with 1 more match to go;
43 wins, 1 draw and 1 defeat, with no matches to go.

15 2139

y – x = 2139.

16 6 The sum of the digits is 9, so all the four-digit numbers that consist of
these digits are divisible by 9. In order for it to be divisible by 36, it also
needs to be divisible by 4. In this case the last two digits must be arranged
as follows: ̅̅̅̅̅̅̅ , ̅̅̅̅̅̅̅ , ̅̅̅̅̅̅̅ and ̅̅̅̅̅̅̅ . It is now possible to find
the four-digit numbers we are looking for: 1260, 2160, 6120, 1620, 2016
and 6012.

17 14 The length of the edge of the cube is 1 cm. The dimensions of the cuboid
are 1 cm ×1 cm ×3 cm , therefore its surface area is

2 × (1×1+1×3+1×3)=14 sq cm.

18 14 1 weaver would weave 2 rugs in 3 days.

3 weavers would weave 6 rugs in 3 days.
3 weavers would weave 2 rugs in 1 day.
3 weavers would weave 14 rugs in 7 days.
19 2 If x = 2, then 5x + 3 = 13 is a prime number.

If x > 2, then it is an odd number; 5x would also be an odd number, and 5x

+ 3 would be an even number, i.e. 5x + 3 would not be a prime number. 
The only prime number is obtained when x = 2.

20 1

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TEAM COMPETITION – NESSEBAR, BULGARIA

MATHEMATICAL RELAY RACE

GROUP 6

Problem 1. The sum of the smallest negative integer, which has an absolute value that is smaller 
than 4, and the smallest positive integer, which has an absolute value that is not smaller than 4, is
@. Find @.

Problem 2. The segment AB has a length of @ meters. The point C separates it in a 1:4 ratio, 
starting from point A. The point D is the midpoint of the segment AC, and the point E is between 
the points D and B, and divides the segment DВ in a 1:4 ratio. The distance from point E to point C 
is # cm. Find #.

Problem 3. Less than 80 passengers were traveling on a bus. Half of them had occupied the seated 
spaces. # % of all passengers got off at the first stop. The number of passengers that got off the bus 
is &. Find &.

Problem 4. At least how many numbers do we need to choose among all the numbers from (-100) 
to 100, in order to be sure that at least 18 of them are divisible by & without leaving a remainder? 
Denote the answer using §. Find §.

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 @ = 1 The numbers are respectively – 3 and 4. Their sum is 1.

2 # = 8

𝑐𝑚 𝑐𝑚
𝑐𝑚 𝑐𝑚

3 & = 4

The number of passengers who got off is of all passengers, i.e. a
number that is divisible by 25. Out of the the natural numbers smaller
than 80, only 25, 50 and 75 are divisible by 25. However, seeing as half
of the passengers were seated, then their number must be even.

Therefore the number of passengers is 50. Four passengers got off at the
first stop.

4 § = 168

The numbers from 100 to 100 are 201. Among them, 25 positive
numbers, 25 negative numbers, and the number 0 are divisible by 4. A
total of 51.

In order to make sure that we have chosen at least 18 which are divisible
by 4, we would need to choose 201 51 +18 = 168 numbers.

5 * =14 Let us reach А in х ways. In this case we would be able to reach С in 12

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MATHEMATICS WITHOUT BORDERS 
2015-2016

AUTUMN 2015: GROUP 7

Problem 1. If

, then & really represents the operation:

A) + B) C) D)

Problem 2. The value of the expression is:

A) 98,406 B) 9,846 C) D) 60,489

Problem 3. The sum of the absolute values of all integers x satisfying | | and | | > 3 is:

A) 0 B) 4 C) 6 D) 8

Problem 4 . What is the largest natural number n for which

A) 7 B) 8 C) 9 D) 10

Problem 5. If

and

then is

A) B) – C) 0 D) 1

Problem 6. Points A, B and C lie on a straight line. The distance from A to B is 16 cm. The 
distance from C to A is 10 cm. Find the distance from the midpoint of BC to the midpoint of AB.

A) 3 cm B) 5 cm C) 6 сm D) 8 сm

Problem 7. A vessel contains 44 litres of water, while a second vessel contains 8 litres of water. 
The same amount of water has been added to both vessels so that the water in one of the vessels
has become 4 times the amount of water in the other vessel. How many litres of water has been
added to each container?

A) 8 B) 5 C) 4 D) 36

Problem 8. If then equals

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А) 9 B) 10 C) 11 D) 12

Problem 10. If the absolute values of the numbers x and y are equal, and , the
expression

| | | |

|

takes:

А) 1 value B) 2 different values C) 3 different values D) 4 different values 
Задача 11. If

then the value of is….

Problem 12. Find the value of А, if

Problem 13. In a group of 80 people, 39 have brown hair, 30 have brown eyes and 15 have both 
brown hair and brown eyes. How many have neither brown hair nor brown eyes?

Problem 14. There are X points marked on a circumference, 5 of which are red, and the rest are 
blue. Each two of the points are connected by a segment. If the number of segments with two red
ends is equal to the number of segments with different-colored ends, how much is X?

Problem 15. The natural numbers are grouped as follows:

What is the sum of the numbers in the tenth group?

Problem 16. I added 100 g mixture of milk and cocoa in a ratio of 4 : 1 to 150 g mixture of milk 
and cocoa in a ratio of 1 : 4. What is the ratio of the milk and cocoa in the newly received
mixture?

Problem 17. What is the three-digit number for which the equality is true

̅̅̅̅̅ ̅̅̅ ̅̅̅ ̅̅̅ ?

Problem 18. How many are the five-digit numbers which end in 6 and are divisible by 3?

Problem 19. Find the smallest prime number, which can be represented as a sum four different 
prime numbers.

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 C

2 A 
3 D |x|<5 and |x|>3⇒x=4 or x=-4

|4|+|-4|=8

4 B

5 D ⇒

⇒

6 B If the point C is between A and B, then the distance would be 5cm. 
If the point A is between B and C, then the distance would be 5cm.
Keep in mind that it is NOT possible for B to be between A and C.
7 C Carry a check using the given answers: 44 + 4 = 48 and 8 + 4 = 12;

48 12 = 4.

Or use the following solution:

Pour an extra x litres of water. Then 44 + x = 4 (8 + x) ⇒ x = 4

8 A

9 D Other than checking through the given answers, the problem can be 
solved as follows:

If we mark the number we are looking for with an x, then 
⇒ х = 12. Possible answer.

10 C The possibilities are –3, –1 and 1. This is how we calculate them:
1 + 1 – 1 = 1; 1 – 1 – 1 = –1; –1 + 1–1 = –1; –1–1–1 =–3

11 ⇒

12 1 First way: we substitute it with x = 1 and conclude that 
0 = 10 – 15 + А + 8 – 5 + 1

Second way:

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13 26 With brown hair but not with brown eyes: 39 – 15 = 24;
With brown eyes, but not with brown hair: 30 – 15 = 15.
With brown eyes and with brown hair: 15.

Deduct 24 + 15 + 15 = 54 from the total number in order to get the
number of people who have neither brown eyes nor brown hair:
80 – 54 = 26.

14 7 The number of segments with two red ends is 10.

The number of segments with two different colored ends is 5(X-5)
От ⇒

15 1729

Their sum is
Another way: 82 + 83 + ... + 99 + 100 = 1729.

The Indian mathematician Ramanujan claims that the number 1729

is the smallest natural number that can be presented, in two

different ways, as the sum of two different numbers raised to the
third power.

16 11:14 The milk in the first mixture is 30g, and the cacao is 120g. In the 
second mixture the milk is 80g, and the cacao is 20g. The total
amount of milk is 110g, and the total amount of cacao is 140g.
The ratio is now 11:14.

17 198 ⇒
⇒

18 3000 The five digit numbers that end in 6 would be divisible by 3 if the 
four-digit number that we can get after removing the digit of ones of
the five digit number is divisible by 3. The total number of four digit
numbers are

9999 – 999 = 9000. Each third number is divisible by 3.
19 17 2 + 3 + 5 + 7 = 17

20 16

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WINTER 2016: GROUP 7 
Problem 1.

A) B) C) D) other

Problem 2. In 1808 the German mathematician Carl Gauss introduced the square bracket
notation to denote the largest integer not greater than .

What is the value of the following expression: ?

A) -2 B) -1 C) 0 D) 1

Problem 3. The value of the expression

is equal to:

A) 1,000 B) 10,000 C) 100,000 D) 1,000,000

Problem 4. One of the interior angles of a triangle is 70 degrees, and the difference of two of its 
interior angles is 30 degrees. How many triangles with such angles are there?

(Hint: The sum of the interior angles of a triangle is 180 degrees.)

A) 1 B) 2 C) 3 D) 4

Problem 5. You can see that 4 points have been placed on the square grid. How many acute 
angles are formed by intersecting the straight lines that connect each pair of points?

(Hint: When two straight lines intersect at a point, they form either 2 acute and 2 obtuse angles or 4 right angles.)

A) 4 B) 5 C) 6 D) another answer

Problem 6. What is the remainder after dividing by 15?

A) 0 B) 5 C) 10 D) 15

Problem 7. The product of two integers that are smaller than 7 and greater than ( 77) is 77.
What is the sum of these integers?

A) B) – C) 18 D) 78

Problem 8. When apples are dried they lose 84% of their weight. How many kilos of apples
would be needed to produce 24 kg of dried apples?

A) 84 B) 100 C) 125 D) 150

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A) 10 B) 11 C) 12 D) 13

Problem 10. The code of a safe is made up of all odd digits, without any of the digits repeating. 
What is the maximum number of failed attempts that we would need to make before breaking the
code?

A) 120 B) 99 C) 119 D) another answer

Problem 11. You can see a rectangle with a size of 11 9 cm. What is the maximum number of
smaller rectangles with sizes of 3 2 cm we can form out of the big rectangle?

Problem 12. The natural number A would be increased 11 times if, on its right side, we write 
down one of the following nine digits: 1, 2, 3, 4, 5, 6, 7, 8 or 9. How many digits does the
number A have?

Problem 13. In how many different ways can we divide a set of 7 different weights (from 1 to 7

grams each) in two groups of equal weight?

Problem 14. How many proper irreducible fractions are there, whose numerator and denominator
are natural numbers with a sum of 33?

Problem 15. What is the smallest natural number N for which the product of 13, 17 and N can be 
presented as the product of three consecutive natural numbers?

Problem 16. Someone asked Pythagoras about the time, and his reply was as follows:

“The time left until the end of the day is equal to two times two fifths of the time which has
already passed”. (twenty-four-hour period). What is the time?

Problem 17. If , the value of the expression would be ( 1). What would the
value of the expression be if ?

Problem 18. How many prime numbers „ ” are there, for which all three numbers
and are primes?

Problem 19. There were 18 apples in one basket, and 20 apples in another basket. I took a few 
apples from the first basket, and then I took as many apples as were left in the first basket, from
the second basket. How many apples in total are left in both baskets?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 B

2 A [–20.15] + [20.15] + [–20.16] + [20.16] = –21 + 20 – 21 + 20 = –2

3 D

4 B Since the sum of the angles of a triangle is equal to two right angles = 180
degrees, the triangles that satisfy the condition of the problem are two and their
interior angles are of the following degrees:

70, 40, 70 and 70, 10, 100.

5 D Let А, В, С and D denote the points so that D is not on the same line as А, В 
and С.

There are 6 pairs of lines that connect each pair of points: AD and AC, AD and
BD, AD and CD, AC and BD, AC and DC, BD and DC.

When two straight lines intersect at a point, they form either 2 acute and 2 
obtuse angles or 4 right angles. AD and BD are perpendicular, hence they form
4 right angles. Each of the rest 5 pairs of lines form 2 acute and 2 obtuse
angles, hence the total number of acute angles is 10.

6 C quotient + remainder

Hence the remainder is a natural number, that is a multiple of 5, but is smaller
than 15. i.e. it is either 0, 5 or 10.

The remainder is not 0, because cannot be divided by 3.

The remainder is not 5, because in this case

The number equal to = 1 ⏟

is not divisible by 3.

Therefore the remainder is 10.

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Among these, two integers are smaller than 7 and greater than 77, hence the
numbers we are looking for are 7 and 11. Their sum is 18.

8 D 100 % 84 % = 16 %
16% of x is 24 kg

⇒

We need 150 kg of apples to produce 24 kg dried apples.

9 C А = 2n + 1 + 2n + 3 + 2n + 5 = 6n + 9,

therefore B = 2n – 3 + 2n – 1 + 2n + 1 = 6n – 3 
or B = 2n + 5 + 2n + 7 + 2n + 9 = 6n + 21. 
Therefore A – B = 12 or A – B = –12.

10 C All possible codes are 5 4 3 2 1 = 120. In the worst case scenario, we
would break the code after making 119 failed attempts. The code would be the
120th possible number.

Therefore we would break the code after the 119th attempt.

11 D 99 6=16 + a remainder of 4, therefore 16 is most likely to be the greatest
number.

12 1 If we write down the number ̅̅̅̅ where x is one of the digits from 1 to 9, then 
10A + x = 11A. Then x = A, i.e. A is a one-digit number.

13 4 The total weight of the set of weights is 28 grams. If the weight of 7 grams is
in the first group, then we would need to add other weights in the group that
have a total weight of 7 grams. 7 = 6 + 1 = 2 + 5 = 3 + 4 = 1 + 2 + 4, therefore
we could group them in 4 ways.

First way: first group – 7, 6, 1; second group – 2, 3, 4, 5

Second way: first group – 7, 5, 2; second group – 1, 3, 4, 6

Third way: first group – 7, 4, 3; second group – 1, 2, 5, 6

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14 10 First way: We write down all proper fractions with 33 as the sum of their
numerator and denominator and then we remove all reducible fractions: 1/32;
2/31, 4/29, 5/28; 7/26, 8/25; 10/23; 13/30; 14/19; 16/17.

Second way: The numerators of the fractions we are looking for would be all
natural numbers, smaller than 17 and non divisible by divisors of 33 other than
1 and 33, i.e. 3 and 11.

15 600 13 4 17 3 = 52 51, therefore the smallest natural number we are
looking for is 4 3 50 = 600.

13 17 600 = 50 51 5
16 13 h 20

min

The time that has already passed can be denoted with x. The time left can be 
denoted with 2 . Therefore the equation can be presented as

⇒ ⇒
17 therefore A = 2.

Then if x = 2, the value of the expression would be

18 1 If p is a prime number, then p = 3, or when divided by 3 leaves a remainder of 
1 or 2.

If p = 3, all three numbers are prime: 3, 14, 17.

If, when divided by 3, the number p leaves a remainder of 1, then the number

p+14 is divisible by itself, by 1, and by 3, i.e. it is not a prime number.

If, when divided by 3, the number p leaves a remainder of 2, then the number

p+10 is divisible by itself, by 1, and by 3, i.e. it is not a prime number.

Therefore there is only one number (p = 3) for which all three numbers 
and are primes.

The total number of apples left is 18 x + 2 + x = 20.

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SPRING 2016: GROUP 7

Problem 1. If , then the value of the expression | | | | is:

A) +1 B) C) D)

Problem 2. What would need to be equal to, in order to get the smallest possible value of the
following expression ?

A) 2016 B) 1 C) D) 2

Problem 3. The sum of the two-digit numbers ̅̅̅ and ̅̅̅ cannot be …

A) 66 B) 154 C) 198 D) 155

Problem 4. The value of the expression

can be calculated after simplifying the expression .
The value of is:

A) 2017 B) 2016 C) 2015 D) other

Problem 5. All possible values of the parameter , for which 3 is the root of the equation
, are the numbers:

A) 3 B) C) and D) and

Problem 6. To what power do we need to raise in order to get ?

A) 2 B) 6 C) 12 D) 24

Problem 7. If and , then is:

A) a non-negative number B) a negative number

C) a positive number D) cannot be determined

Problem 8. If each of the angles of a quadrangle is the average of the other three angles, then
this quadrangle would always be:

A) a parallelogram B) a rhombus C) a square D) a rectangle

Problem 9. How many of the solutions of the equation are solutions of
the inequality | |

A) 0 B) 1 C) 2 D) 3

Problem 10. After a price increase of 20 % an item cost $ 240. The price of the same item before 
the price increase was:

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Problem 11. The numbers 0, 1, 2, 3, 4 and 5 have been used to write down all four-digit
numbers that have no repeated digits and are divisible by 5. What part of these numbers are
divisible by 10?

Problem 12. 26 litres of juice must be bottled into 10 bottles of either 1 litre, 3 litres or 5 litres. 
The 1 litre bottles are of an even number. How many bottles of 5 litres are there?

Problem 13. If | | | | where is a positive integer, what is the smallest
value of

Problem 14. If is a prime number, how many possible remainders would we get when dividing

p by 6?

Problem 15. The polynomial is expressed as follows:

What is the value of ?

Problem 16. On each side of a cube with an edge of 1 cm is glued a cube equal in size. Find the 
surface area of the resulting body in cm2.

Problem 17. 12– 22 + 32– 42 + 52– 62 +...+ 992 −1002 = …

Problem 18. In a chess tournament, each contestant must play a game with each of the other
contestants. There are 4 contestants in the tournament: Alexander, Boris, Carl and David. So far
Alexander has played 3 games, Boris has played 2 games and Carl has played 2 games. How

many games has David played?

Problem 19. If the product of 100 numbers is a negative number, how many different options are 
there for the possible number of negative integers among these multipliers?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 D ⇒ | |
For each | | .

Therefore:

| | | |

2 D 
=

We can get an equality when both and . We get
that .

3 D ̅̅̅ ̅̅̅
Therefore the sum of ̅̅̅ ̅̅̅ is always divisible by 11.
Among the given numbers 154 is not divisible by 11.

4 D ⇒

5 D If is a root, then е ⇒ ⇒
⇒ ⇒ or

6 C Let denote the number we are looking for.

⇒ ⇒ ⇒ ⇒

7 B and ⇒

and ⇒

and ⇒ ⇒ ⇒

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⇒ ⇒

In the same way we can get that

Thus we reach the conclusion that the quadrangle is a rectangle.

9 C The numbers 2, 3 and – 3 are solutions of the equation

Among them 3 and – 3 satisfy the inequality | |

10 A Let denote the price we are looking for
In this case

⇒

11 5/9 The numbers divisible by 5 are either of the ̅̅̅̅̅̅̅ kind or of the

̅̅̅̅̅̅̅ The numbers of the ̅̅̅̅̅̅̅ kind are divisible by 10.

The numbers of the ̅̅̅̅̅̅̅ kind are 5 4 3 = 60, and the numbers of the

̅̅̅̅̅̅̅ kind are 4 4 3 = 48.

The part we are looking for is 60 108 = 5/9.

12 Let x, y and z denote the number of bottles containing 1, 3 and 5 liters. 
Then ⇒ ⏟

⇒ ⇒
We have these options to bottle the juice:
If z = 1 ⇒ y = 6 ⇒ x = 3;

If z = 2 ⇒ y = 4 ⇒ x = 4; 
If z = 3 ⇒ y = 2 ⇒ x = 5.

Since the 1 litre bottles are of an even number, only the second option is
possible – the number of 5 litre bottles is 2.

13 If N = 1 ⇒ | | | |

If N = 2 ⇒ | | | | .

If ⇒ | | | |

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possibilities would be and .
However in and the number p is not a 
prime. Then the possible remainders are 1 and 5.

Now we must add the remainders 2 and 3, which we get when dividing the
primes 2 and 3 by 6.

If ⇒ the possible remainders are 1 and 5;
If ⇒ the possible remainders are 2 and 3.
There are 4 possible remainders in total.

15 6 The identity = 
is also true for . Therefore

= ⇒ .
16 30 The number of cubes is 7. The resulting body is made up of 5 faces of

each of the 6 cubes = 30 faces in total, each one with an area of 1 sq.cm 
= 30 sq.cm.

17 – – – –

= ⏟

18 1 or 3 If we add the number of games each contestant has played, the number we
get must be divisible by 2, because each game is counted twice.

In this case the number of games is 7 + x, where x denote the number of 
games David has played. The number x cannot be greater than 3.

There are four possibilities: 0, 1, 2 and 3.
7 + x can be divisible by 2 if x = 1 or x = 3.

19 50 The negative numbers must be an odd number. The possibilities are 1, 3,
5, ..., 99. 50 in total.

20 2 Let denote the price of an apple. The price of a pear would then be
= .

From the second condition we get that the price of a pear is equal to

We get the equality ⇒
Then = 0.5.

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FINAL 2016: GROUP 7

Problem 1. The product of 100 integers is 100. What is the smallest possible sum of these 
numbers?

A) -199 B) -195 C) -2 D) 0

Problem 2. The numbers from 0 to 100 have been written down in order: 
01234567891011...979899100. If we were to erase three consecutive digits, the first two of
which have a sum of 10, the third digit will most often be:

A) 0 B) 1 C) 2 D) 3

Problem 3. The sum of the two-digit numbers ̅̅̅ and ̅̅̅ is a square of a natural number and is
equal to:

A) 196 B) 169 C) 144 D) 121

Problem 4. What remainder is left after dividing by 12?

A) 0 B) 2 C) 4 D) 8

Problem 5. If the following is true for each value of a:

then = ?

A) 2 B) 1 C) 2 D) 4

Problem 6. At least how many composite integers, smaller than 50, do we need to select, so that 
at least two of them would have a common divisor that is greater than 1?

A) 4 B) 5 C) 6 D) 7

Problem 7. The number (-1) is the root of the equation

where is the unknown and A is a parameter. 
In this case A =

A) -1 B) 0 C) 1 D) 2

Problem 8. The point M is placed on the side BC of the triangle ABC so that CM = MA = AB 
and AC = BC. In this case =

A) 2 B) 1.5 C) 2.5 D) 3

Problem 9.

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Problem 10. We have a square with a side of 10 cm. We have cut out smaller squares, each with 
a side of 1cm, from two opposite corners of the big square. What is the greatest possible number 
of rectangles with sizes of 1cm by 2 cm that we can divide the newly formed figure into?

A) 98 B) 49 C) 48 D) different answer

Problem 11. Determine all integers A, such that = ̅̅̅̅ and the two-digit number ̅̅̅̅ can
be presented as the product of two consecutive odd numbers?

Problem 12. The water content in 5 kg of fresh mushrooms is 90%. After drying the mushrooms, 
the water content is now 20% of their weight. How many grams do the dried mushrooms weigh? 
Problem 13. In a particular year there are three consecutive months with 4 Sundays in each. 
What are the possible sums of the numbers of days in those three consecutive months?

Problem 14. Let us have a look at the following pairs of numbers: 
.

Find n, if the sum of the digits of the numbers of each pair is 11.

Problem 15. An isosceles triangle with a leg of 2 cm and a square with a side of 1 cm have equal 
areas. How many degrees is the smallest angle of the triangle?

Problem 16. If N is an integer, how many are the possible remainders when dividing by 5?
Problem 17. After having covered of my journey, plus another 200 m, I have yet to go 
another 50 m less than the distance I have already covered. How many km is the length of my 
journey?

Problem 18. In 1808, the German mathematician Carl Gauss introduced the indication .
He used it to denote the greatest integer that is not greater than .

Calculate the value of the expression [[
]

] [ ].

Problem 19. Five weavers weave 10 rugs in 3 days. How many rugs would 3 weavers weave in 
7 days?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 A

100 = ( 100) ⏟

⇒

the sum we are looking for is ( 199)

2 С The digit triplets are 910, 192, 282, 829, 373, 738 , 464, 647, 555, 556, 646,
465, 737, 374, 828, 283, 919, 192 .The third digit is most often 2.

3 D

̅̅̅ ̅̅̅
⇒

⇒ ̅̅̅ ̅̅̅

4 С

, therefore it is not hard to conclude

that the remainders are among the numbers 0, 2, 4, 6, 8 and 10. It is not
possible for the remainder to be 0 or 6, because 3 can not divide
It is not possible for the remainder to be 2, because 4 divides but can
not divide 12 quotient + 2.

There are two possibilities left: 4 or 8.

The remainder is 4, because only is divisible by 3, but
is not divisible by 3.

5 D

therefore

=

6 В

Each composite integer that is smaller than 50 is divisible by one of the prime
numbers 2, 3, 5 and 7.

Therefore according to the principle of Dirichlet, it would be enough to
choose 5 numbers in order to be sure that among them there would be two
which have a common divisor greater than 1.

7 D ⇒
⇒

8 A ⇒
⇒

9 A

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10 C

Let us paint the board using a chess pattern. The squares which have been cut
out are of the same color, and each of the rectangles 1 2 covers both colors.
In this case it would not be possible to get 49 rectangles. However it is
possible to get 48.

(Two of the squares can not be used to form a rectangle – the ones adjacent to
the squares which have been cut out)

11 2

Out of the perfect squares 16, 25, 36, 49, 64 and 81, only 36 satisfies the
initial condition: 63 can be presented as the product of two consecutive odd
numbers.

In this case the number A can either be 6 or -6.

12 625

Let us denote the weight of the mushrooms in kilograms using x.

We can compare the weight of the “dry” material:

⇒ .

13 89 or 90

The possibilities are:

31+28+31=90 или 31+29+31=91;
28+31+30=89; 29 + 31 + 30=90.
31+30+31=92;
30+31+30=91;
31+30+31=92;
30+31+31=92;
31+31+30=92;
31+30+31=92;
30+31+30=91;
31+30+31=92.

Between 12 Sundays there are 12 + 11 6 = 78 days;

Therefore from the sums we will have the following options left:
12 days or 13 days; 11 or 12; 14; 13; 14; 14; 14; 14; 13; 14. 
February is definitely among those months.

The possibilities are:

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14 28

The numbers smaller than 28 do not satisfy the condition of the problem. For
example, is not possible, one of the groups will have the numbers 19
and 9, and in this case the sum of the digits would be 19.
The numbers greater than 29 do not satisfy the condition of the problem
either, because there will be a group that has the number 29 and the number
. However the sum of the digits of the number is a number
greater than 0. In this case in the group of numbers the sum of
the digits would be greater than 11.

15 15 or 30

If the triangle has an acute angle between its legs, then it would be . In
this case, the smallest angle would be - that would be the angle between
the legs.

If the triangle has an obtuse angle between its legs then it would be . In
this case, the smallest angle would be – these would be the angles at the
base.

16 2 The fourth power of any natural number has 0, 1, 5 or 6 as its digit of ones. In
this case when divided by 5 the remainder would be either 0 or 1.

17

⇒

Let us denote my full journey with X m. In this case 
⇒

18 [[ ]] [

]

19 14

1 weaver will weave 2 rugs in 3 days;
3 weavers will weave 6 rugs in 3 days;
3 weavers will weave 2 rugs in 1 day;
3 weavers will weave 14 rugs in 7 days.

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TEAM COMPETITION – NESSEBAR, BULGARIA 
MATHEMATICAL RELAY RACE

The answers to each problem are hidden behind the symbols @, #, &, § and * and are used in 
solving the following problem. Each team, consisting of three students of the same age group,
must solve the problems in 45 minutes and then fill a common answer sheet.

GROUP 7

Problem 1. is an irreducible fraction with natural numbers for a numerator and a denominator,
such that and The number of all such fractions is @. Find @.

Problem 2. The number @ has shown up on a computer screen. After each hour that passes, the 
number gets replaced by the sum of the number and the product of its digits. Which number will
show up on the screen after 2016 hours? Denote the answer using #. Find #.

(If the number is 11, the number that shows up on the screen after the first hour will be equal to

11 + 1 1 = 12, then the number that shows up on the screen after the second hour will be 
equal to

12 + 1 2 = 14, ....)

Problem 3. Two squares have sides that are measured in integer centimeters. The sum of the 
areas of the squares is (# + 68) . The greatest possible perimeter of the figure is & cm. 
Find &.

Problem 4. The product of & consecutive two-digit numbers is divisible by . The smallest
number in this product is §. Find §.

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 @ = 24

The numerator a can be any integer from 1 to 52. In order for the fraction 
to be irreducible, we must eliminate all of the possible numerators that are
divisible by 3, 5 and 7 (the prime divisors of 105).

The numbers divisible by 3 are 17.
The numbers divisible by 5 are 10.
The numbers divisible by 7 are 7.

The numbers divisible by 3, and by 5, are 3.
The numbers divisible by 3, and by 7, are 2.

The number divisible by 5, and by 7, is 1.

Therefore the numbers that are divisible either by 3, or by 5, or by 7, are 17
+ 10 + 7 – (3 + 2 + 1) = 28.

The irreducible fractions are 52 – 28 = 24.

2 # = 170

The numbers that will show up on the screen are:
24; 24 + 2 4 = 32; 32 + 3 2 = 38; 38 + 3 8 = 72;

72 + 2 7 = 86; 86 + 8 6 = 134; 134 + 1 3 4 = 146; 146 + 1 4 6
= 170; 170 + 1 7 0 = 170; 170; 170; ...

3 & = 58

If the lengths x and y are the lengths of the sides of the squares, the 
perimeter of the figure is 4x + 2y, x > y.

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4 § = 18

The product of the numbers from 1 to 67 contains

[ ] [ ] numbers 5 when factorized as a product of
simple multipliers.

The product of the numbers from 1 to 9 contains

[ ] [ ] number 5 when factorized as a product of simple
multipliers.

In this case the product of the numbers from 10 to 67 contains 15 - 1 = 14
numbers 5 when factorized as a product of simple multipliers.

10 11 .... 66 67 contains 14 multipliers 5;
11 12 …. 67 68 contains 13 multipliers 5;
12 13 …. 68 69 contains 13 multipliers 5;
13 14 .... 69 70 contains 14 multipliers 5;
14 15 .... 70 71 contains 14 multipliers 5;
15 16 .... 71 72 contains 14 multipliers 5;
16 17 …. 72 73 contains 13 multipliers 5;
17 18 …. 73 74 contains 13 multipliers 5;
18 19 .... 74 75 contains 15 multipliers 5.
The smallest number we are looking for is 18.

5 * = 6

Let the sides be denoted as a, b and c and let . In this case
⇒

From the inequality of the sides of the triangle ⇒
 ⇒ we can find the possible values of c: 6 cm, 7 cm and 8 cm. 
If ⇒ .

If ⇒ .

If ⇒

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MATHEMATICS WITHOUT BORDERS 
2015-2016

AUTUMN 2015: GROUP 8
Задача 1. If then the value of is:

A) 2,015 B) C) 4,030 D)

Problem 2. When the natural number is divided by 6, the remainder is 2. When the natural number
is divided by 6, the remander is 3. What is the remainder when is divided by 6?

A) 0 B) C) 4 D) 5

Задача 3. If and , then | |

A) 3 B) C) 4 D)

Problem 4. Find the tens digit in the value of

⏟

A) 0. B) 1 C) 2 D) 5

Problem 5. If ab>0 and a+b<0, then the value of | | | | is:

А) B) C) D) 0

Problem 6. A polygon has more than 40 diagonals. The number of its sides is at least:

А) 9 B) 10 C) 11 D) 12

Problem 7. When solving the same quadratic equation, three students got different roots: 
The first student got the numbers 1 and 2 as roots;

The second student - 2 and 3;
The third student - 3 and 4.

It turned out that each student got exactly one root of the equation right.
If и are the roots of the equation, then – is:

A) 1 B) 2 C) 4 D) 9

Задача 8. If √ and √ , how many of the numbers and 
are rational?

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Problem 9. The numbers a and b are such that the expression has the smallest
value possible. The value is:

A) B) 1 C) 2 D)

Problem 10. A rectangle is divided by two intersecting lines parallel to its sides. Four smaller 
rectangles are formed, three of which have areas of 3 sq.cm, 4 sq.cm and 5 sq.cm. Find the smallest

possible area of the fourth rectangle.

А) B) C) D)

Problem 11. What is the largest possible number of acute angles in a convex hexagon? 
Problem 12. How many are the five-digit numbers which end in 6 and are divisible by 3?

Problem 13. How many are the integers smaller than 2015, which can be represented as a sum of
two consecutive integers and as a sum of three consecutive integers?

Problem 14. For how many integers , the value of is a prime number?

Problem 15. What is the greatest number of cells 1x1 you can paint in a 11x11 square drawn on
graph paper, so that no one box 2x2 did not have three of painted cells?

Problem 16. For how many integers , the numbers

and

are also integers?

Problem 17. In how many ways can we distribute 7 identical pears between 3 children so that each
child receives at least one pear?

Problem 18. Find the integer , if √ √ √ .

Problem 19. The point D is of the median CM of the triangle ABC and it is such that СD=DM. If the 
point E is an intersection of the straight line AD and the side BC, determine CE:CB.

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 C 
⇒ X=8060, Y=4030⇒X-Y=4030

2 A

3 A ⇒ ⇒ | |

4 A After we cancel the fraction, we would get a number that is the
product of a number and 100. The digit of tens is 0.

⏟

5 B ⇒ | |
| | ⇒

⇒ | | | |

6 C The number of diagonals of an N-triangle can be found using the 
formula In this case, if we carry out a check, we will find that:

.

Therefore, the polygon would have 11 angles.

7 C The roots are either 1 and 3, or 2 and 4. Therefore the value we are
looking for is 4.

8 C From √ и √ , it follows 
that only and are rational numbers.

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10 D If we mark the areas of the 4 rectangles with and ,
.

Then the smallest area would be

11 3 The sum of the angles of a convex hexagon is 720◦ and each of them is
smaller than 180◦. If 4 of the angles are acute, then the sum of the
other two would be greater than 360◦, i.e. at least one of them would
be greater than 180◦.

12 3000 The five digit numbers ending in 6 are divisible by 3 if:

The four digit number that we got after removing the digit of ones
from the five digit number, is divisible by 3. The total number of four
digit numbers is 9999-999=9000. Each third number is divisible by 3.
13 336 The sum of two consecutive natural numbers is odd, and the sum of

three consecutive natural numbers is divisible by 3.

The odd numbers that are smaller than 2015 and are divisible by 3

are 3, 9, 15, ..., 2013. Their total number is 336.

14 3 The numbers

17-0, 17-4 are the primes for three of the values of n=3, 5, 1

15 66 If we paint all odd columns, the number of colored squares would be

⏟

16 0

;

⇒ ⇒ ,

which is impossible.

17 15 First solution: 7=5+1+1=4+2+1=3+3+1=3+2+2, therefore the

following positioning can be done in:
5, 1, 1 - 3 ways;

4, 2, 1 – 6 ways;
3, 3, 1 – 3 ways
3, 2, 2 – 3 ways.
15 ways in total.

Second solution: Let us add two identical apples to the pears and

arrange the 8 pieces of fruit in a row.
Let us now arrange the pears as follows:

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On the second row – from the first to the second apple;
On the third row – from the second apple to the end.

We can place the first apple in the second place, in which case the
places available for the second apple would be 4 to 8, i.e. 5
possibilities.

We can place the first apple in the third place, in which case the
places available for the second apple would be 5 to 8, i.e. 4
possibilities.

We can place the first apple in the fourth place, in which case the
places available for the second apple would be 6 to 8, i.e. 3
possibilities.

We can place the first apple in the fifth place, in which case the places
available for the second apple would be 7 to 8, i.e. 2 possibilities.
We can place the first apple in the sixth place, in which case there
would only be one place left available for the second apple, that is, 8,
i.e. 1 possibility.

5+4+3+2+1=15 possibilities in total.

18 2 First solution: √ √ ⇒ √

√ ⇒ ⇒ .

Second solution: Apply the following formula

√ √ √ √ √ √ .

19 1:3 ⇒ 
⇒

M is the midpoint of AB. Therefore the areas of and are
equal ⇒

Since the areas of the triangles ACM and BCM are equal, it follows
that 2F=3S+F⇒F=3S⇒ 4S=1:2.

⇒

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WINTER 2016: GROUP 8

Problem 1. If the equation is an identity, then

A) B) C) D) other

Problem 2. The square of the natural number A is recorded with the digits 0, 2, 3 and 4. Which digits 
would we use to record 5 A?

A) 0, 2, 3 B) 0, 2, 4 C) 2, 3 and 4 D)another answer

Problem 3. If , then
| |

A) 17 B) 33 C) 65 D) 129

Problem 4. One of the interior angles of a triangle is 70 degrees, and the difference of two of the
interior angles of this same triangle is 30 degrees. How many such triangles are there?

(Hint: The sum of the interior angles of a triangle is 180 degrees.)

A) 0 B) 1 C) 2 D) 3

Problem 5. You can see that 4 points have been placed on the square grid. How many obtuse angles
will be formed by intersecting the straight lines that connect each pair of points?

(Hint: When two straight lines intersect at a point, they form either 2 acute and 2 obtuse angles or 4 
right angles.)

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Problem 6. It is known that the sum of more than 2 consecutive natural numbers is 20. How many 
possibilities are there?

A) 0 B) 1 C) 2 D) 3

Problem 7. Find the area (in sq cm) of an isosceles triangle with an angle of and a 10 cm long 
leg.

A) 100 B) 50 C) 25 D) 12.5

Problem 8. Someone asked Pythagoras about the time, and his reply was as follows:

“The time left until the end of the day is equal to two times two fifths of the time which has already
passed”. (twenty-four-hour period) What is the time?

A) 13 h 20 min B) 13 h 40 min C) 14 h 20 min D) 14 h 40 min

Problem 9. Peter added 3 consecutive odd numbers and the sum he got as a result was A. Steven 
added 3 consecutive odd numbers and the sum he got as a result was B. If one of Peter’s numbers is
the same as one of Steven’s numbers, then the greatest possible difference of the sums they both got
(A and B), is:

A) 10 B) 11 C) 12 D) 13

Problem 10. In a rectangular triangle, and are the legs, is the hypotenuse, is the height to the
hypotenuse. Which of the following sums is the greatest?

A) B) C) D)

Problem 11. You can see a rectangle with a size of . What is the maximum number of
smaller rectangles with sizes of we can form out of the big rectangle?

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Problem 13. The isosceles triangles and have been built on the sides of the square
If , then calculate

Problem 14. How many proper irreducible fractions are there, whose numerator and denominator are
natural numbers with a sum of 41?

Problem 15. What is the smallest natural number N, for which the product of 13, 17 and N can be 
presented as the product of three consecutive natural numbers?

Problem 16. Calculate √ √ √ √ .

Problem 17. Find the perimeter of the quadrilateral that we would get when connecting the midpoints
of the sides of a quadrilateral with diagonals equal to 4 cm and 5 cm.

Problem 18. Find the sum of the coefficients of the even degrees in the simple form of the
polynomial

Problem 19. What is the greatest possible value of the number N that would make the following 
statement true: “N numbers can always be found among 97 random integers, in a way that the 
difference of each pair would be divisible by 8.”?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 C First way:

Hence = 125 + ( 300) + 240 + ( 64) = 1.

Second way:

If x = 1, then ⇒
2 B The square of a natural number can not end neither in 2, nor in 3. If it ends in

0, that means that it would end in two zeros. Therefore the square of number

A ends in 4. The possibilities are 3204; 3024; 2304 and 2034.

We would have to exclude 2034 as a possibility because A is an even number 
and the square of A is divisible by 4, whereas 2034 is not divisible by 4.

From 3204 = 4 9 89, 3024 = 4 9 84; 2304 = 4 9 64 it follows that

number A is 48, and 5 48 = 240.

3 A 
, therefore

√ , i.e. | |

4 C Since the sum of the angles of a triangle is equal to two right angles = 180
degrees, the triangles that satisfy the condition of the problem are two and
their interior angles are of the following degrees:

70, 40, 70 and 70, 10, 100.

5 D Let А, В, С and D denote the points so that D is not on the same line as А, В 
and С.

There are 6 pairs of lines that connect each pair of points: AD and AC, AD and

BD, AD and CD, AC and BD, AC and DC, BD and DC.

When two straight lines intersect at a point, they form either 2 acute and 2 
obtuse angles or 4 right angles. AD and BD are perpendicular, hence they 
form 4 right angles. Each of the rest 5 pairs of lines form 2 acute and 2 obtuse
angles, hence the total number of obtuse angles is 10.

6 В There is 1 possibility: 2 + 3 + 4 + 5 + 6 20.

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Therefore the height to the leg would be 5 cm and the area would be 25 sq.cm.

АВС is an isosceles triangle ⇒ АС = ВС

Let АН⏊BC (H∊BC) ⇒Δ ACH, HCA= ⇒ ⇒
the area of = 25 sq cm

8 A The time that has already passed can be denoted with x. The time left can be 
denoted with 2 . Therefore the equation can be presented as

⇒ ⇒
9 C

therefore – – –
or
Therefore – или – –

10 D From and

it follows that is the greatest sum.

11 16 99 6 = 16 + a remainder of 4, therefore 16 is most likely to be the greatest
number.

12 240 Let us mark the two students as student A and student B. We can arrange
them and the other four students in 5 4 3 2 1 = 120 ways. In each of
these ways the students A and B can be arranged either as AB or as BA.
Therefore they can be arranged in a total of 2 120 = 240 ways.
13 is equilateral, therefore ANM =

is isosceles ⇒ ⇒

ΔABM is isosceles with legs AM and BM and MAB = .

Therefore Δ ABM is equilateral and the measure we are looking for is 60 
degrees.

14 20 First way: We write down all proper fractions with 41 as a sum of their
numerator and denominator and then we remove all reducible fractions, if
there are any. All proper fractions with 41 as a sum of their numerator and
denominator are irreducible: 1/40, 2/39, 3/38,..., 20/21.

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than 1 and 41. The possible numerators are all natural numbers from 1 to 20.
15 600 13 4 17 3 = 52 51, therefore the smallest natural number we are

looking for is 4 3 50 = 600.
13 17 600 = 50 51 5

16

√ √ √ √ √ √ √ √

| √ | | √ | ( √ ) √

17 9 cm The quadrangle we get as a result is а parallelogram with sides equal to the
halves of the diagonals, i.e. 2cm; 2.5cm; 2cm and 2.5cm. The perimeter is 9

cm.

18 1

⇒

19 13 The difference of two numbers would be divisible by 8 if they both leave the
same remainder when divided by 8.

The number of remainders after a division by 8 is 8: 0, 1, 2, 3, 4, 5, 6 and 7.
Among the 97 numbers that are not divisible by 8:

12 leave a remainder of 0;
12 leave a remainder of 1,
12 leave a remainder of ... ,
……….. ,
12 leave a remainder of 7.

12 therefore there is 1 number left. This number will also leave a

remainder which will be a number between 0 and 7.

Therefore the greatest number of pairs of numbers among 97, with a
difference divisible by 8 is 13.

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SPRING 2016: GROUP 8

Problem 1. The greatest negative integer that is a solution of the inequality | | √ is:

A) B) C) D)

Problem 2. The sum of three numbers is 222. If we increase the first number by 2, increase the
second number twice, and diminish the third number twice, we would get the same number. Find the
smallest of the three numbers.

A) 22 B) 32 C) 42 D) other

Problem 3. What is the value of the following expression?
√ √ √

A) √ B) √ C) D) 1

Problem 4. What power do we need to raise to, in order to get ?

A) 2 B) 6 C) 12 D) 24

Problem 5. There are 8 points on the circumference of a circle. What is the greatest possible number 
of right-angled triangles that have these points as their vertices?

A) 24 B) 30 C) 36 D) 4

Problem 6. Two years ago A was twice older than B, and three years ago B was three times younger 
than A. How old is A now?

A) 12 B) 10 C) 8 D) 6

Problem 7. For how many of the integers n can we claim that is divisible by ?

A) 0 B) 1 C) 2 D) more than 2

Problem 8. What remainder is left when is divided by 13?

A) 6 B) 4 C) 2 D) 0

Problem 9. 26 litres of juice must be bottled into 10 bottles of either 1 litre, 3 litres or 5 litres. In how
many ways can we do this, if we use all three bottle sizes?

A) 2 B) 3 C) 4 D) 7

Problem 10. In the graph | |, where is the parameter, and the coordinate axes determine a
triangle with an area of . What is the smallest possible value of the expression ?

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Problem 11. If N and M are natural numbers such that √ √ , calculate .

Problem 12. (inspired by a problem by Johannes Buteo, who lived during the 16th century) The price 
of 9 apples, minus the price of a pear, is equal to $13, and the price of 15 pears, minus the price of an
apple, is equal to $ 6. How much would one apple and one pear cost?

Problem 13. The diagonals of a trapezium divide it into four triangles. Three of the areas of the
triangles are equal to respectively 4, 6 and 9 sq.cm. What is the area of the trapezium?

Problem 14. What is the number of real roots of the equation | | ?

Problem 15. If A and B are 4-digit natural numbers, how many solutions can you find to the 
following equation ?

Problem 16. The numbers 187 and 219 leave the same remainder (11) when divided by the number
. Find the number .

Problem 17. Four children: A, B, C and D must be arranged in a row, in such a way that A and B, and

C and D, would always be standing next to each other. In how many different ways can we do this?

Problem 18. Calculate the sum of the reciprocal and the opposite of number if √

Problem 19. If each of the angles of a quadrangle is the average of the remaining three angles, find
the largest angle.

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 B Check each possible answer, starting with and until you
reach the answer, that is

2 B If denote the number that we get after increasing the first number by
2, then the first number would be , the second number would be ,
and the third number would be . As a result, we would get the

following equation:

The first number is 62, the second number is 32 and the third number is
128.

3 C

√ √ ( √ ) | √ | ( √ ) (√ ) ( √ )

4 B If denote the number we are looking for ⇒ ⇒

⇒ ⇒ ⇒

5 А If we place the points two by two, so that they would be the edges of a
diameter, we would get 6 right-angled triangles with a common
hypotenuse for each diameter. There are 4 right-angled
triangles in total.

6 D А B

Three years
ago

Two years ago

The eqation is

⇒

Now

7 B

and therefore

is an integer if

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8 D

The expression is divisible by 13, and the remainder would be 0.

9 B Let and denote the number of bottles of 1 l and 3 l respectively 
Then the bottles of 5 litres would be – – .

⇒
By using we can fill in the following table:

1 l 3 l 5 l

6 bottles 0 bottles 4 bottles

5 bottles 2 bottles 3 bottles

4 bottles 4 bottles 2 bottles

3 bottles 6 bottles 1 bottles

2 bottles 8 bottles 0 bottles

From the table above we can see that the number we are looking for is 3.

10 B In the graph of | |, where is the parameter, and the
coordinate axes determine the triangle

with an area of ⇒
.

In this case the smallest possible value of the expression is 0.

11 3 √ √ √

If ⇒ √ is an irrational number, then .
Therefore M=1 and M+N=3.

12 2 Let x denote the price of an apple. The price of a pear would then be 
equal to 9 - 13.

From the second condition we get that the price of a pear is equal to

We get the equality

⇒

Then = 0.5.

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13 25 ABCD is the trapezium, O is the intersection of its diagonals, AB .
The areas of the triangles ADO and BCO are equal, therefore the 
possible areas of the four triangles are 4, 4, 6, 9; 4, 6, 6, 9; 4, 6, 9,9.

and and ,

therefore the areas of the triangles are 4, 6, 6 and 9.

In this case the area of the trapezium is 6 + 6 + 9 + 4 = 25.

14 2 | | ⇒

, if . The root is equal to 0. 
 , if . The root is equal to

15 6984 , therefore the value of A is the smallest, if B 
The value of A is the greatest, if .

The numbers are 7983 – 999 = 6984.

16 16 We are looking for a natural number greater than 11, which is a divisor
of both 187 – 11 and 219 – 11. This number is 16.

17 8 We must arrange the X and Y pairs, which are respectively made up of

A and B, C and D.

X and Y can be arranged in 2 ways, and each of X and Y can also be 
arranged in 2 ways. 2 2 2 = 8 ways in total.

18 2

√ ⇒ √ √ ⇒

19 90 The sum of the angles of every quadrangle is 360 degrees. Let
the angles

⇒ ⇒ ⇒ ⇒
In the same way we can reach the conclusion that
t

20 30 The identity = 
is also true for . Therefore

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FINAL 2016: GROUP 8

Problem 1. If the number a is rational and the number √ is also rational,
then the smallest value of b is:

A) 0 B) 1 C) 2 D) 3

Problem 2. Take a look at the following pairs of natural numbers
.
If the sum of the digits of the numbers from each pair is 23, find n.

A) 499 B) 994 C) 949 D) different answer

Problem 3. In the numerical equality known as „the problem of the Indian mathematician Bhaskara”
the last number is replaced with the letter A. Find A.

√ √ √ √ √ √

A) 5 B) 6 C) √ D) √

Problem 4. A straight line has been built through the vertex A of the parallelogram ABCD and the 
point M of the diagonal BD and this straight line divides the diagonal BD in a 1:2 ratio starting from 
the vertex D. In what ratio does this straight line divide the side CD starting from the vertex D?

A) 1:1 B) 1:2 C) 2:1 D) 3:1

Problem 5. If the following is true for each value of a,

then

A) - 2 B) -1 C) 2 D) 4

Problem 6. What is the sum of the real roots of the following equation?

A) -1 B) 0 C) 1 D) 2

Problem 7. Find the distance from the intersection point of the graphs of the functions and
to the ordinate axis.

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Problem 8. A square and a circle have a common part. The area of the square, the area of the 
common part and the area of the circle are relative to each other 4:1:17. What percentage of the figure
below is the area of the common part?

A) 5 B) 10 C) 15 D) 20

Problem 9. In 1808, the German mathematician Carl Gauss introduced the indication [ ].
He used it to denote the greatest integer that is not greater than .
How many natural numbers n are there, for which [ ] is a prime?

A) 1 B) 2 C) 3 D) more than 3

Problem 10. How many points (x, y) are there that have negative integers as coordinates and 
?

A) 0 B) 1 C) 2 D) more than 2

Problem 11. is an equilateral triangle with sides of length 3 cm. The points M, N and P are 
respectively found on the sides BA, AC and CB, and are such that MN⏊AC, NP⏊CB and PM⏊AB. 
Calculate the length of the segment AM.

Problem 12. Calculate the difference of the real numbers and , if and .

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Problem 14. What is the smallest possible number of different digits that we can use to write down 6

numbers, which when divided by 6 leave different remainders?

Problem 15. In a particular year three consecutive months have 4 Sundays each. What are the 
possible sums of the days of these three consecutive months?

Problem 16. Calculate the value of the following expression:

√ √ √ √ √ √ √ √ .

Problem 17. We have a square with a side of 10 cm. We have cut out smaller squares, each with a 
side of 1cm, from two opposite corners of the big square. What is the greatest possible number of 
rectangles with sizes of 1cm by 2 cm that we can divide the newly formed figure into?

Problem 18. Calculate the product of the real roots of the following equation:

Problem 19. If N is an integer, how many possible remainders are there when dividing by 5?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 A

The number √ would be a rational number if
. We get two possible values for : 2 and (-1)

In this case the smallest value of is 0.

2 D n = 598

3 C

√ √ √ √ √ √ ⇒

√ √ √ √
⇒ √ √ √ √ ⇒
( √ ) √ √ √ =0⇒ √ .

4 A

In a triangle ACD the median is DO (O is the intersection of the 
diagonals of the parallelogram). However in this case the point M would 
be a centroid, which means that AM will intersect the side CD in its 
midpoint.

5 A

=

i.e.

6 А

First we need to note that 1 is not a root of the equation.
If ⇒ ( )

⇒

We can now come to the conclusion that the roots are 0 and (-1).
Their sum is (-1).

7 B The intersecting point of the two graphs is a point with coordinates of
(3;- 6). The distance we are looking for is 3.

8 A

The areas of the square, the common part, and the circle are respectively
 and The surface of the whole figure is 20k. The percentage we 
are looking for is

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[ ] We get a composite number, because if m = 1, 
then .

If , where m is a natural number or 0, then 
[ ]

We get a composite number because и
If , where m is a natural number or 0, then

[ ] We will only get prime
numbers for i.e. for

10 C The inequality is true only for

11 2

If ⇒

From and the above calculations we can reach the
following equation

Such a triangle does exist.

12 5 or (- 5)

We can conclude that and .
In this case ⇒ | |

The difference between the two numbers is either 5 or (-5).
There are such real numbers.

13 77 283 The greatest product is 76421 853 9

The sum of the multipliers is 76421 + 853 + 9= 77 283.

14 2

For example:

12, 1, 2, 111, 22, 111, when divided by 6, leave the following

remainders: 0, 1, 2, 3, 4 и 5. They can be written down with 2 digits.

15 89 or 90

The possibilities are:

31+28+31=90 или 31+29+31=91;
28+31+30=89; 29 + 31 + 30=90.
31+30+31=92;

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30+31+31=92;
31+31+30=92;
31+30+31=92;
30+31+30=91;
31+30+31=92.

Between 12 Sundays there are 12 + 11 6 = 78 days;

Therefore from the sums we will have the following options left:
12 days or 13 days; 11 or 12; 14; 13; 14; 14; 14; 14; 13; 14.
February is definitely among those months.

The possibilities are: 31+28+31=90; 28+31+30=89; 29 + 31 + 30=90.

16 5 √ √ √ √ √

√ √ √ 
 √ 
 
√ √ 
 
√

17 48

Let us color the board using a chessboard pattern. The squares which
have been cut out are of the same color, and each of the rectangles 1 2
covers both colors.

In this case it would be impossible to get 49 rectangles. However it is
possible to get 48 rectangles.

(Two of the squares can not be used to form a rectangle – the ones
adjacent to the squares which have been cut out)

18 0

The number 0 is a root of the equation, because

In this case the product of the roots is also 0.

19 2

The fourth power of every natural number has 0, 1, 5 or 6 as its digit of
ones. In this case, when divided by 5, the remainder would be 0 or 1.

20 20

Atos did not shake hands with 70 people, Porthos did not shake hands
with 60 people and Aramis did not shake hands with 50 people.

In the worst case scenario, only Atos did not shake hands with those 70
people, Porthos did not shake hands with those 60 and Aramis did not
shake hands with those 50. The total number is 180.

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TEAM COMPETITION – NESSEBAR, BULGARIA

MATHEMATICAL RELAY RACE

GROUP 8

Problem 1. The smallest natural number а, for which the equation 
 || | | has exactly two solutions, is @. Find @.

Problem 2. Find the number # of the natural numbers N, such that √ √ .

Problem 3. In ΔABC the side AB has been divided into 5 equal parts. Straight lines, parallel to AC, 
have been built through the points of division, which has created 4 segments with their ends along the
sides AB and BC of the triangle. If the sum of these segments is # cm, calculate the length of side AC 
in centimeters. Denote the answer using the symbol &. Find &.

Problem 4. The sum of the digits of the number equal to ⏟ is §. Find §.

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 @ = 18

The equation || | | has a solution if
If , the equation has exactly two solutions.

Let We reach the conclusion that

| | or | | . 
| | or | | .

In order for there to be two solutions i.e.
If or the equation will have exactly two solutions.
The value of that we are looking for is the number 18.

2 # =30 √ √ √ √

The number of natural numbers from 5 to 35 is 30.

3 & =15

Let us use to denote the segment that is closest to the vertex B. It is a 
middle segment of a triangle where the next longest segment is 2 , the
one after is 3 and the fourth is 4 . The sum of the four sections is 10 .
We have found that = 3 cm. In this case = 5 = 15 cm.

4 § = 27 ⏟

⏟

The sum we are looking for is 27.

5 * = 3

Consider the following statement:

If among M coins only one is a fake (it is lighter) and M e a random 
number among the numbers and ,

then the minimum number of times we would need to weigh the coins in
order to find the lighter coin is N.

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MATHEMATICS WITHOUT BORDERS 
2015-2016

AUTUMN 2015: GROUP 9

Problem 1. Which of the following is the largest number?

A) √ B) √ C) √ D) √

Problem 2. Find the missing number in the equation

√

A) 25 B) 45 C) 90 D) 135

Problem 3. The number of rational numbers in the sequence √ √ √ √ √ is:

A) 31 B) 16 C) 450 D) 225

Problem 4. In the rectangular triangle , the hypotenuse √ and . The
area of the triangle is:

A) B) √ C) √ D) √

Problem 5. If √ and √ , then the value of | | | | is:

А) B) 2 C) √ D)- √

Problem 6. A polygon has more than 100 diagonals. The number of its sides is at least:

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Problem 7. When solving the same quadratic equation, three students got different roots: 
The first student got the numbers 1 and 2 as roots;

The second student - 2 and 3;
The third student - 3 and 4.

It turned out that each student got exactly one root of the equation right.
If the roots are и , then – is:

A) 1 B) 2 C) 4 D) 9

Problem 8. The numbers a and b are such that the expression 
has the smallest value possible. The value is:

A) B) 1 C) 5 D)

Problem 9. A rectangle is divided by two intersecting lines parallel to its sides. Four smaller 
rectangles are formed, three of which have areas of and . (see the diagram) Find the smallest
possible value of the area of the fourth rectangle if .

А) 𝑆 ×𝑆𝑆 B) 𝑆 ×𝑆𝑆 C) 𝑆 ×𝑆𝑆 D) other

Problem 10. How many are the natural numbers n, for which the number

is also a natural

number?

А) 0 B) 1 C) 2 D) 3

Problem 11. If ( √ ) × (√ ) , what is the maximum value for √ ?

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Problem 13. In trapezoid ) is the length of its leg AD and is the 
distance from the midpoint of AB to the leg AD (see the diagram). If АВ:DC=3:2, express the area of 
the trapezoid in terms of a and b.

Problem 14. How many are the integers smaller than 2015, which can be represented as a sum of four 
consecutive integers?

Problem 15. What is the greatest number of cells 1x1 you can paint in a 11x11 square drawn on graph 
paper, so that no one box 2x2 did not have three of painted cells?

Problem 16. Find the integer , if √ √ √ √

Problem 17. The point D is of the median CM of the triangle ABC and it is such that СD=DM. If the 
point E is an intersection of the straight line AD and the side BC, determine CE:CB.

Problem 18. For which prime numbers and the roots of the equation
are integers?

Problem 19. How many natural numbers are there, that are factors of ?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 D If we compare the numbers

√ √ √ √
we will find that the greatest of them is √ √
2 D the number we are looking for is 135.

3 B The numbers are √ √ √ √ √ . They are as many as the 
odd numbers 1, 3, 5, ..., 29, 31. The odd numbers from 1 to 31 inclusive,
are 16.

4 A Let us assume that С and С are respectively the height and the 
median of the side АВ. In the rectangular С
× √ . The area of the triangle is equal to

×

5 A От √ <0 и √

| | | |
 | | | |

6 B The number of diagonals of a polygon with N angles can be determined
by using the formula × By carrying out a check, we will find that

× . Therefore the polygon has 16 
angles.

7 C The roots are either 1 and 3, or 2 and 4. Therefore the value we are
looking for is 4.

8 B =

×
The conclusion is a classic example:

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9 B If we mark the areas of the 4 rectangles with and ,
× × .

The smallest area would be 𝑆 ×𝑆

𝑆 .

10 C

=3+
is an integer for

The expression is a natural number only for 3 and 7. 
11 1 ( √ ) × (√ ) √ or √

√ or √
√ {√ × √

12 3 The sum of the angles of a convex hexagon is 720◦ and each of them is
smaller than 180◦. If 4 of the angles are acute, then the sum of the other
two would be greater than 360◦, i.e. at least one of them would be
greater than 180◦ .

13

is the midpoint of АВ. The area of the triangle AMD is

DM is a

median in the triangle ABD. Therefore the area of the triangle В D is

The area of the triangle АВD is

The areas of the triangles BCD and АВD are in the same ratio as AB and

CD. Therefore the area of the triangle ABD would be The area of

the trapezium ABCD is .

14 502 The sum of 4 consecutive natural numbers is

when divided by 4 leaves a remainder of 2.

The numbers are 10, 14, 18, ..., 2010, 2014.

(4 × × × ×
502 in total.

15 66 If we paint all odd columns, the number of colored squares would be

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16 -2 First solution:

√ √ и √ √
|√ | |√ | .

Second solution: If we raise to the second power we will get the

following result:

√ √ √ √ √ С
 или .

√ √ √ √ <0

Third solution:

Apply the following formula:

√ √ √ √ √ √ .
17 1:3

is the midpoint of АВ. Therefore the areas of and are 
equal

If the areas of the triangles ACM and BCM are equal, then
2F=3S+F F=3S 4S=1:2.

18 (2;2);

(2;3); 
(3;2)

If we apply Viet’s theorem, we will reach the conclusion that if a and b
are the roots of the equation, then

( ) × ( )

If a>1 and b>1, then the solution is (2;2).

If a=1 or b=1 , then the solutions are (3;2) or (2;3).

19 16 × × , therefore the natural numbers which are
divisors are 16.

20

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WINTER 2016: GROUP 9

Problem 1. After we simplify the following expression and rewrite it in simple form, what would the 
coefficient of be?

A) B) C) D) other

Problem 2. If √ × is a rational number, what is the smallest natural number N?

A) 2 B) 7 C) 14 D) 28

Problem 3. How many different integers can be presented as “x” in the following inequality?

√

A) 1 B) 2 C) 3 D) more than 3

Problem 4. Equilateral triangle ABC is inscribed in a circle. Point M is located on the smaller of the 
two arcs AC. How many angles with vertices A, B, C and M are there on the diagram?

A) 3 B) 4 C) 5 D) 6

Problem 5. The differences of each two of three natural numbers are different. What is the smallest 
possible sum of these three numbers?

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Problem 6. What is the remainder after dividing by 15?

A) 0 B) 5 C) 10 D) 15

Problem 7. If and are the real roots of the equation , then | | is equal to:

A) 5 B) √ C) √ D)

Problem 8. The integers from 1 to 50 have been written down on 50 cards. What is the minimum 
number of cards that we would need to draw without looking in order to make sure we have drawn at

least 3 cards with prime numbers on them?

A) 38 B) 39 C) 40 D) 50

Problem 9. An isosceles triangle has a perimeter of 56 cm, and two of its sides have a ratio of 3:2. 
What is the shortest possible side of the triangle?

A) 10 cm B) 14 cm C) 16 cm D) 24 cm

Problem 10. How many natural numbers that contain three different digits, one of which is a 
geometric mean of the other two, are there?

(Hint: If × , a is the geometric mean of b and c.)

A) 6 B) 12 C) 18 D) 24

Problem 11. If the numbers a, b ∊ √ √ √ , how many ordered pairs of numbers (a,

b) are there, where either a + b, or a × b are natural numbers?

Problem 12. What is the ones digit of the number equal to the sum of the cubes of the numbers from 1 
to 101?

Problem 13. If ABCD is a square, and the point F is such that the triangle BCF is equilateral, what is 
the greatest possible measure that AFD can have?

Problem 14. How many proper irreducible fractions are there,whose numerator and denominator are
natural numbers with a sum of 14?

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Problem 16. The angles of the vertices B and C of the triangle ABC are respectively and . The

point M is interior to the triangle and Calculate

Problem 17. Calculate A if

Problem 18. Find the values of the parameter a, where the coordinate axes and the straight lines 
and form a trapezium.

Problem 19. How many integers from 2000 to 2016 are there, that can not be values of the 
discriminant of the quadratic equation with integer coefficients?

Problem 20. On the diagram below you can see anisosceles triangle ABC with an area of and
legs AC and BC, each with a length of 10 cm. Point M is on the base AB, and points E and F are feet of 
altitudes from point M to the legs BC and AC. Find the greatest possible value of the product ME×

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 А In the simple form of the polynomial

× ,

we can get the ninth degree from

× × × × ×

The coefficient is 0.

2 C If √ × is a rational number, then the number equal to × × ×
must be a perfect square.

Therefore the exponents must be even numbers and this is possible if N is at 
least 2×7.

× × × × × × × × ×

3 B

√ ∊

4 C Just like the angles of the triangle ABC, the angles AMB and BMC are also 
equal to 60 degrees.

5 B Let us assume that the numbers are 1, 2 and n. In this case the differences are 
 . The smallest value of n is 4. Therefore the sum we are 
looking for is 1 + 2 + 4 = 7.

6 C × quotient + remainder

Hence the remainder is a natural number, that is a multiple of 5, but is smaller
than 15. i.e. it is either 0, 5 or 10.

The remainder is not 0, because cannot be divided by 3.

The remainder is not 5, because in this case × ×
The number equal to × = 1 ⏟

is not divisible by 3.

Therefore the remainder is 10.

7 B ×

The equation does not have real roots, because its discriminant
3 < 0.

Then и are real roots of the equation .

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8 А There are 15 primes among the numbers from 1 to 50. Those numbers are 2, 3,
5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. In this case we would need to
take out at least 35+3=38 cards in order to make sure that there are at least 3
primes among them.

9 B There are two triangles that can satisfy the condition of the problem:
A triangle with sides of 21, 21, 14; A triangle with sides of 16, 16, 24.
10 D We are looking for the numbers containing the digits (1, 2, 4), (1, 3, 9), (2, 4,

8) and (4, 6, 9).

Their total number is 24:
124; 142; 214; 241; 412; 421;
139; 193; 391; 319; 913; 931;
248; 284; 482; 428; 824; 842
469; 496; 649; 694; 946; 964
11 4 The numbers are

( √ √ ) (√ √ ) ( √ √ ) ( √ √ )

12 1 The cubes of ten consecutive numbers have a ones digit equal to 1, 8, 7, 4, 5, 6,
3, 2, 9, 0.

The ones digit of the sum of these cubes is 5.

The sum of 100 consecutive numbers will end with the same digit as 10 × 5,
i.e. 0.

The sum of the numbers from 1 to 101 will end with the same digit as
10 × 5 + 1, i.e. 1.

13 150 If point F is external to the square, then the angle we are trying to find would 
be 30 degrees, and if it is interior to the square, the angle would be 150
degrees.

14 3 First way: We write down all proper fractions with 14 as a sum of their
numerator and denominator and then we remove all reducible fractions.

The fractions are 1/13, 3/11 and 5/9.

Second way: The numerators of the fractions we are looking for would be all
natural numbers, smaller than 7 and non divisible by divisors of 14 other than 1
and 14, i.e. 2 and 7. The possible numerators would be 1, 3 and 5.

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15 600 13 × 4 × 17 × 3 = 52 × 51, therefore the smallest natural number we are
looking for is 4 × 3 × 50 = 600.

13 × 17 × 600 = 50 × 51 × 5

16 140 Point M is the center of the circle that encircles the triangle. The angle we are

looking for is central and

17 1,008

( × )

×

( × ) ×

18 0 and 1 The straight lines and the coordinate axes would form a trapezium if the
straight lines are parallel to one another. In this case a = 1.

A trapezium would also be formed when a = 0, in which case the straight line 
 would be parallel to the x axis.

19 8 If the equation is , then
If the number b is even, then D would be divisible by 4.

If the number b is odd, then after dividing D by 4, there would be a remainder 
of 1.

In this case the numbers among the integers from 2000 to 2016, that are not
values of the discriminant, are: 2002, 2003, 2006, 2007, 2010, 2011, 2014,
2015.

20 4 If x = MF and y = ME × × × × 
We are looking for the greatest value of the expression × ×

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SPRING 2016: GROUP 9

Problem 1. If × √ , then √

A) B) C) D)

Problem 2. The numbers 2 and are two of the four roots of the equation
What is the sum of the other two roots?

A) B) 0 C) 1 D) other

Problem 3. What is the value of the following expression? 
√ √ √

A) √ B) √ C) 1 D) 1

Problem 4. How many solutions does the following equation have? 
× √

A) B) C) 2 D) 3

Problem 5. There are 8 points on the circumference of a circle. What is the greatest possible number 
of right-angled triangles that have these points as their vertices?

A) 24 B) 30 C) 36 D) 4

Problem 6. Two years ago A was twice older than B, and three years ago B was three times younger 
than A. How old is A now?

A) 12 B) 10 C) 8 D) 6

Problem 7. For how many of the integers n can we claim that is divisible by ?

A) 0 B) 1 C) 2 D) more than 2

Problem 8. What remainder is left when is divided by 21?

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Problem 9. The circle inscribed in the right-angled triangle ABC touches the hypotenuse AB at point

M. If the radius of the circle is 1 cm, and AM = 3 cm, then AB – BC =

A) B) C) D)

Problem 10. In the graph | |, where is the parameter, and the coordinate axes determine a
triangle with an area of . What is the smallest possible value of the expression ?

A) B) 0 C) 8 D) other

Problem 11. If N and M are natural numbers, such that √ √ calculate N

Problem 12. The diagonals of a trapezium divide it into four triangles. Three of the areas of the 
triangles are equal to respectively 4, 6 and 9 sq.cm. What is the area of the trapezium?

Problem 13. What is the number of real roots of the equation | | ?

Problem 14. Six children, A, B, C, D, E and F, must stand in a row in such a way that A and B, C and 
D, E and F would always be standing next to one another. In how many different ways can we do this
arrangement?

Problem 15. The polynomial is written in the following form:
× × What is the value of ?

Problem 16. The numbers 201 and 235 leave the same remainder (14), when divided by . Find .
Problem 17. If

√ √ √ √ , find A.

Problem 18. The number of diagonals of a convex N-gon is 2015. What is the number N?

Hint:√

Problem 19. What would the last digit (the units digit) of the square of an integer be, if the digit 
before-last (the tens digit) is odd?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 C (√ ) (√ ) (√ ) √ √

2 C First way: We solve the equation to find that the
roots are equal to 2, 2 , 3 and 3. The sum we are looking for is 1.

Second way: All equations of the kind that have and
as solutions also have and – as solutions. In this case the other
two solutions are equal to 3 and 2. Their sum is equal to 1.

3 D √ √ √ | √ | √ ( √ ) √

4 B × √ or √ the possible 
roots are 1, 1 or 2. The check shows that only 2 is a root of the equation.

5 А If we place the points two by two, so that they would be the edges of a
diameter, we would get 6 right-angled triangles with a common

hypotenuse for each diameter. There are 4 × right-angled triangles
in total.

6 D А B

3 years ago

2 years ago

The equation is

Now

7 B

and therefore

is an integer,

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8 D

× ×

× ×

Тhe remainder after a division by 21 is 1.

9 B If AB = c, BC = a, CA = b

10 B In the graph of | |, where is the parameter, and the coordinate
axes determine the triangle with an
area of .

In this case the smallest possible value of the expression is 0.

11 3 √ √ √

If √ is an irrational number. Therefore .
We can now reach the conclusion that M =1. Therefore M+N =3.

12 25 ABCD is the trapezium, O is the intersection of its diagonals, AB .
The areas of the triangles ADO and BCO are equal, therefore the possible 
areas of the four triangles are 4, 4, 6, 9; 4, 6, 6, 9; 4, 6, 9, 9.

and 𝑆𝑆 and 𝑆𝑆 ,

therefore the areas of the triangles are 4, 6, 6 and 9.
In this case the area of the trapezium is 6 + 6 + 9 + 4 = 25.

13 2 | |

, if . The roots are the numbers 0 and 1.

, if . This equation has no real roots

14 48 We must arrange the X, Y and Z pairs, which are respectively made up of

A and B, C and D, E and F.

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15 13 The identity = × × 
is also true for . Therefore

= × × .

16 17 We need a natural number greater than 14, which is a divisor of both
and . That number is 17.

17 2

√ √ √ √

√ √

√ √
( √ √ ) √ √

√ √

√

√ √

18 65 The number of diagonals is determined by the following formula:

19 6 , therefore the digit before last would
be odd if the tens digit of is odd. This would be possible if b or
The ones digit would be 6.

20 72 The lengths of the medians of the triangle ABC are, respectively, 9

cm, 12 cm and 15 cm. The point M is the centroid of the triangle. 
Let us duplicate the triangle and get the parallelogram АСВD, where the 
point N is the centroid of the triangle АВD.

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FINAL 2016: GROUP 9

Problem 1. If √ √ √ , then √

A) B) C) D)

Problem 2. A circle has been inscribed in the right-angled triangle ABC. The circle touches the 
hypotenuse AB at the point M. If AM = 3 cm and BM = 6 cm, then the area of the triangle is:

A) B) C) D)

Problem 3. If the number a is rational and the number √ is also rational,
then the value of b is:

A) 0 B) 1 C) 2 D) 3

Problem 4. What is the product of the real roots of the following equation?

A) B) C) D)

Problem 5. Of all the triangles with sides a, b and c, such that ,
calculate the perimeter of the triangle with the largest area.

A) 25 B) 24 C) 23 D) other answer

Problem 6. Find the natural number , 97 < < 102, for which the expression
 is a perfect square of a natural number.

A) 98 B) 99 C) 100 D) 101

Problem 7. If the points M and N are the midpoints of the sides CD and DA of the parallelogram

ABCD, and the straight lines AM and BN intersect at the point P, then the =

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Problem 8. A square and a circle have a common part. The area of the square, the area of the common 
part and area the circle relate to each other in a ratio of 4:1:17. What percentage of the figure's area is
the area of the common part?

A) 5 B) 10 C) 15 D) 20

Problem 9. The number of rational numbers in the sequence √ √ √ √ √ is:

A) 44 B) 42 C) 22 D) 21

Problem 10. How many points (x, y) are there, with coordinates that are negative integers and 
?

A) 0 B) 1 C) 2 D) more than 2

Problem 11. The equation , where a and b are parameters, has a double 
root 1. How many real roots does it have?

Problem 12. The acute-angled triangle ABC is inscribed in a circle with a center O and a radius of R. 
If r is the radius of the circle tangent to the segments AO and BO, and the arc AB, and 
calculate .

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Problem 14. Present the following as an irreducible fraction 
̅ ̅
̅ ̅

Problem 15. The sides of the triangle ABC are: AB = 3 cm, BC = 4 cm and AC = 5 cm. Points K, M 
and N are the feet of the perpendiculars from point P to the sides AB, AC and BC respectively. 
Calculate 3AK+4ВN+5СM.

Hint: One of the classic theorems in geometry is that of the French mathematician Lazare Nicolas

Carnot: The perpendiculars drawn from points K, M and N to the sides AB, AC and BC of the triangle

ABC intersect at point P only when

Problem 16. If , and , calculate

Problem 17. In the numerical equation √ √ √ √ √ √ , known as 
“problem of the Indian mathematician Bhaskara” the last number has been replaced with the letter A. 
Find .

Problem 18. In a particular year three consecutive months had 4 Sundays in them. What are the 
possible sums of the days in these three consecutive months?

Problem 19. We are given a square with a side of 10 cm. If we were to cut out smaller squares, each 
with a side of 1cm, from two of the opposite corners of the big square, what is the greatest possible
number of rectangles with a size of 1 cm by 2 cm that we would be able to cut the newly formed figure
into?

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 B √ √ √ √ √

2 A

If and are the lengths of the sides of the triangle
and is its semiperimeter,

×

3 D The number √ would be rational if
In this case

4 B

First we need to note that 1 is not a root of the equation.
If ( )

It is no longer difficult to establish that the roots are 0 and (-1). Their
product is 0.

5 B

Of all the triangles with sides a and b, the right-angled triangle with 
catheti and , and a hypotenuse equal to √ ∊
has the largest area.

The parameter is equal to 24 cm.

6 A

therefore if , then the expression would be a square of the
natural number, i.e. .

We must note that if then is such that
 ;

We must also note that there is no perfect square between the squares of
two consecutive natural numbers.

7 A

Let us assume that the straight line intersects the continuation of the side

CD at the point Q.

From the congruence of the triangles NDQ and NAB we get that

DQ = AB.

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8 A

The area of the square, the common part, and the circle are respectively
 and The area of the whole figure is 20k. The percentage we are 
looking for is

9 C

The numbers are

There are 22 numbers. We must also note that 
10 C The inequality is true for

11 2

The equation does not have any real roots
has only 2 real roots.

12 30

and are tangent to the point D, аnd Е is tangent to 
 and AO.

In this case In the triangle the cathetus
is twice as small as the hypotenuse . We can conclude

that

13 17

Therefore

In this case .
There are 17 integers of the interval .

14

15 25 If

16

From the condition of the problem we get that и
.

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17 √

√ √ √ √ √ √

√ √ √ √
√ √ √ √
( √ ) √ √ √ =0 √ .

18 89 or 90

The possibilities are:

31+28+31=90 or 31+29+31=91;
28+31+30=89; 29 + 31 + 30=90.
31+30+31=92;

30+31+30=91;
31+30+31=92;
30+31+31=92;
31+31+30=92;
31+30+31=92;
30+31+30=91;
31+30+31=92.

Between 12 Sundays there are 12 + 11 × 6 = 78 days;
Therefore the following will remain of the sums

12 days or 13 days; 11 or 12; 14; 13; 14; 14; 14; 14; 13; 14.
February is definitely among those months.

19 48

Let us colour the board using a chessboard pattern. The blocks which have
been cut off are of the same color, and each of the rectangles 1х2 covers
both colors.

In this case it would be impossible to get 49 rectangles. However it is
possible to get 48.

(Two of the squares can not be used to form a rectangle – the ones
adjacent to the squares which have been cut out.)

20 2

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TEAM COMPETITION – NESSEBAR, BULGARIA

MATHEMATICAL RELAY RACE

GROUP 9

Problem 1. The smallest integer that can take on the parameter a, and for which the equation 
( √ ) × √ is only satisfied for one number, is @. Find @.

Problem 2. The height to the hypotenuse of a right-angled triangle is @ cm. The smallest possible 
area of this triangle is # . Find #.

Problem 3. After an awarding ceremony, the three winners were greeted by a total of # people. The 
first was greeted by 80 people, the second was greeted by 60, and the third by 70. What is the smallest
possible number of people that greeted all three winners? Denote the answer using the symbol &. Find 
&.

Problem 4. The number of all non-negative integers k, for which the inequality 
 √ has a solution presented as a positive integer, is §. Find §.

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ANSWERS AND SHORT SOLUTIONS

Problem Answer Solution

1 @ = 10.

The equation can have a maximum of 2 roots, a and √ .

In order for there to be 1 root, it would be enough for √ , or
√ , i.e. √ .

The value would be equal to 10.

2 # = 100

The median of the hypotenuse is always at least 10 cm. In this case the 
hypotenuse would be at least 20 cm.

Therefore the smallest possible area of this triangle would be .

3 & =10

The first was not greeted by 20 people, the second was not greeted by 40
and the third was not greeted by 30. In the worst case scenario:

Those who did not greeted the first winner did not greeted at least one of
the other two either.

Those who did greeted the second winner did not greeted at least one of
the other two. Those who did not greeted the third winner, did not

greeted at least one of the other two either. In this case 20 + 40 + 30 = 90
people did not greeted the three winners. The people who greeted all
three would be at least 100 90 = 10.

4 § = 2

The smallest possible k, for which the inequality √
has a solution, is 0 and the greatest is 1.

The number we are looking for is 2.

4 * = 2/3

If A, B and C are the centers of the three circumferences, then AB = 2,

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Đề thi toán học không biên giới năm 2016

<b>? = </b>

<b> ? = </b>

+ +

<sub> </sub>

.

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