I.v.

In the multiplication example written on the board, the hooligan Petya changed two digits. It became 4$\cdot$5$\cdot$4$\cdot$5$\cdot$4 = 2247. Restore the original example.

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In the original example, at least one of the multipliers is even.

## Solution

In the resulting example, three multipliers are even, which means that at least one was even in the original example. Therefore, the product was an even number, meaning the last digit of the product was changed. Thus, no more than one digit on the left was changed. Therefore, in the original example, there were both fives and fours on the left, and the product ended in 0.

![](https://cdn.mathpix.com/cropped/2024_05_06_99ef25effe2e32f29eccg-44.jpg?height=63&width=1906&top_left_y=2770&top_left_x=36)
The number 1600 differs from the number 2240 by more than one digit. Therefore, exactly one of the multipliers was corrected, and the product is 2240. Therefore, one of the fives was corrected to a seven.

![](https://cdn.mathpix.com/cropped/2024_05_06_99ef25effe2e32f29eccg-45.jpg?height=305&width=2035&top_left_y=202&top_left_x=0)

Author: Cherapanov.

Let $M$ be a finite set of numbers. It is known that among any three of its elements, there are two whose sum belongs to $M$.

What is the maximum number of elements that can be in $M$?

## Hint

Consider either the four largest or the four smallest numbers.

## Solution

An example of a set with 7 elements: $\{-3, -2, -1, 0, 1, 2, 3\}$.

We will prove that a set $M = \{a_1, a_2, \ldots, a_n\}$ with $n > 7$ numbers does not have the required property. We can assume that $a_1 > a_2 > a_3 > \ldots > a_n$ and $a_4 > 0$ (changing the signs of all elements does not change our property). Then $a_1 + a_2 > a_1 + a_3 > a_1 + a_4 > a_1$, which means that none of the sums $a_1 + a_2$, $a_1 + a_3$, and $a_1 + a_4$ belong to the set $M$. Moreover, the sums $a_2 + a_3$ and $a_2 + a_4$ cannot both belong to $M$, since $a_2 + a_3 > a_2 + a_4 > a_2$. Therefore, for at least one of the triples $(a_1, a_2, a_3)$ and $(a_1, a_2, a_4)$, the sum of any two of its elements does not belong to the set $M$.

## Answer

7 elements.