Task 5. In how many ways can a pile of 100 stones be divided into piles so that the number of stones in any two piles differs by no more than one?
Task 5. In how many ways can a pile of 100 stones be divided into piles so that the number of stones in any two piles differs by no more than one?

## Solutions and Evaluation Criteria

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| 1 | Answer: No. Suppose this is possible. No digit is 0, because then the product on one side is 0, while there is no such equality on the other side. Therefore, all 8 letters are digits from 1 to 9, meaning they include all digits except one. Among them, there is a 5 or 7. But then one of the products is divisible by 5 (7), while the other is not. Contradiction. | 7 points for the correct solution. 1 point for proving that none of the digits is 0. 2 points for proving that there is no 0 and 5-2 points. |
| 2 | Answer: No. Color the vertices in white and black so that the colors alternate. Then the difference between the sum of the numbers in the white vertices and the sum of the numbers in the black vertices is initially 1, and this difference does not change with any move (1 is added to both sums). Suppose the opposite. Then both sums are divisible by 3, because all numbers are divisible by 3. Therefore, their difference is also divisible by 3, but 1 is not divisible by 3. Contradiction. | 7 points for the correct solution, otherwise 0 points. |
| 3 | Answer: $70^{\circ}$. Solution: Triangle $ABD$ is equilateral, so the angles $\angle ABD = \angle BDA = \angle DAB = 60^{\circ}$. Let $\angle BAC = \angle CBD = \alpha$, then $\angle ABC = 60^{\circ} + \alpha$. Since $AB = AC$, $\angle ACB = \angle ABC = 60^{\circ} + \alpha$. The sum of the angles in triangle $ABC$ is $\alpha + (60^{\circ} + \alpha) + (60^{\circ} + \alpha) = 180^{\circ}$, so $3\alpha = 60^{\circ}$, $\alpha = 20^{\circ}$. The angle $\angle CAD = \angle BAD - \angle BAC = 60^{\circ} - 20^{\circ} = 40^{\circ}$. Since $AC = AD$, $\angle ACD = \angle ADC = \beta$, the sum of the angles in triangle $ACD$ is $40^{\circ} + 2\beta = 180^{\circ}$, so $2\beta = 140^{\circ}$, $\angle ACD = \beta = 70^{\circ}$. | 7 points for the correct solution. 4 points for finding the angle $\angle BAC = \angle CBD = 20^{\circ}$. |
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| 4 | Answer: $x = y = 0$. Solution: If both numbers are not 0, divide the numbers $x$ and $y$ by their greatest common divisor, resulting in coprime numbers $a$ and $b$. The right side of the equation is divisible by 3, so the left side must also be. The square of an integer can give a remainder of 0 or 1 when divided by 3, so $a^2$ and $b^2$ are divisible by 3, meaning $a$ and $b$ are divisible by 3. Contradiction, because $a$ and $b$ are coprime and have a common divisor 3, which is greater than 1. | 7 points for the correct solution. 1 point for the correct answer only. 3 points for noticing that $x$ and $y$ are divisible by 3. |
| 5 | Answer: 99. Solution: We will prove that for any $k$ from 2 to 100, the pile can be divided into $k$ such piles in a unique way. $100 = qk + r$, where $q$ and $r$ are the quotient and remainder of 100 when divided by $k$, respectively. Suppose there is a pile with no more than $q-1$ stones, then in the remaining $k-1$ piles, there are at least $101 - q = q(k-1) + (r+1)$ stones. Then in some pile, there are at least $q+1$ stones, otherwise, the number of stones in $k-1$ piles would not exceed $q(k-1)$, which is less than $q(k-1) + (r+1) = 101 - q$. Two piles have been found that do not meet the condition (one with at least $q+1$ and the other with no more than $q-1$). Contradiction. Suppose there is a pile with at least $q+2$ stones. Then in the remaining $k-1$ piles, there are no more than $99 - q$ stones. Then in some pile, there are no more than $q$ stones, otherwise, the number of stones in $k-1$ piles would be at least $(k-1)(q+1) = (kq + k) - 1 - | For the correct solution - 7 points. For the correct answer "99" - 1 point. For the answer "100" - also 1 point. |
|  | $q > (kq + r) - 1 - q > 100 - 1 - q = 99 - q$. Two piles have been found where the number of stones differs by at least 2. Then in all piles, there are $q$ or $q+1$ stones. Let the number of piles with $q$ stones be $t$, then the number of piles with $q+1$ stones is $k-t$, the total number of stones is $100 = qt + (q+1)(k-t) = qt + qk - qt + k - t = (qk + r) - r + k - t = 100 - r + k - t$, so $t = k - r$, i.e., the way is unique. |
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