# 53.

## Solution

We will consider that the week starts on the day of the week on which January 1st of the year in question falls (i.e., if January 1st is a Wednesday, then we consider the week to start on Wednesday and end on Tuesday). Then, from the beginning of the year, [365/7] = 52 (in a leap year [366/7] = 52) full weeks pass, each of which has exactly one Friday. In addition, at the end of the year, there are 365 - 7 * 52 = 1 or 366 - 7 * 52 = 2 days left. One of these remaining days can be a Friday. In such a year, the maximum possible number of Fridays - 53 - is achieved.

## Task 2

Is it possible to place 4 ones, 3 twos, and 3 threes in the circles of a five-pointed star (see the figure on the right) so that the sums of the four numbers on each of the five lines are equal?

![](https://cdn.mathpix.com/cropped/2024_05_06_3c3766a13b29d22de154g-1.jpg?height=194&width=205&top_left_y=1910&top_left_x=1348)

Number of points 7

## Answer

No, it is not possible.

## Solution

Suppose the numbers can be placed in the required manner. Let $S$ be the sum of the numbers on each line. Then the sum of the numbers on all five lines is 5S. Since each circle lies at the intersection of two lines, each number in a circle is counted twice. Therefore, the sum we found is twice the sum of all the placed numbers, i.e., $5S = 2(4 \cdot 1 + 3 \cdot 2 + 3 \cdot 3)$. We get $5S = 38$, which is impossible because on each line there are 4 integers and their sum must be an integer.

## Task 3

Integers $a$ and $b$ are such that $96a = 73b$. Prove that $a + b$ is a composite number.

## Number of points 7

## Solution

If $a$ and $b$ have a common factor, then $a + b$ is composite. Now suppose $a$ and $b$ are coprime. Then $73(a + b) = 73a + 73b = 73a + 96a = 169a$. Since the numbers 73 and 169 are coprime, $a + b$ is divisible by 169. Since 169 is a composite number, $a + b$ is also composite.

## Task 4

Does there exist a hexagon that can be cut into four triangles by a single straight line?

## Number of points 7

Answer

Yes, it does exist.

## Solution

In the figure on the right, a hexagon is shown that can be cut into four triangles.

![](https://cdn.mathpix.com/cropped/2024_05_06_3c3766a13b29d22de154g-2.jpg?height=391&width=503&top_left_y=935&top_left_x=1302)

## Task 5

Matilda gave Pinocchio the task: "Count the ink blots in your notebook, add 7 to their number, divide by 8, multiply by 6, and subtract 9. If you do everything correctly, you will get a prime number." Pinocchio got everything mixed up. He counted the ink blots accurately, but then multiplied their number by 7, subtracted 8 from the result, divided by 6, and added 9. What answer did Pinocchio get?

## Number of points 7

## Answer

Pinocchio got $18 \frac{1}{6}$.
# Solution

Consider the calculations according to Malvina's plan. Adjacent operations "divide by 8" and "multiply by 6" can be replaced by a single multiplication operation by $6 / 8 = 3 / 4$. If after multiplying by $3 / 4$ we obtained a fractional number, then subtracting 9 from it would again result in a fractional number, but we should get a prime (i.e., an integer) number. Therefore, the result of multiplying by $3 / 4$ must be an integer. Moreover, this number will be divisible by 3 (since the 3 in the numerator of $3 / 4$ cannot be canceled out). Then, after subtracting 9, we get a number that is also divisible by 3. And it is known that this number is prime. The only prime number divisible by 3 is 3 itself. Therefore, according to Malvina's plan, the final result should be 3. By performing the operations in reverse order, we find the number of blots:

$$
(3+9): 6 \cdot 8-7=9
$$

Therefore, Buratino obtained the number:

$$
(9 \cdot 7-8): 6+9=18 \frac{1}{6}
$$