Problem 4. On the board, the expression is written

$$
7 * 6 * 5 * 4 * 3 * 2 * 1
$$

Masha replaces the asterisks with the signs «+» and «-». In one move, Vanya can change two consecutive signs to their opposites. Vanya wants to achieve, after several of his moves, an expression whose value is divisible by 7. Can Masha arrange the signs in such a way that Vanya cannot achieve his goal?
Solution. Answer: cannot.

We will show how Vanya can achieve an expression divisible by 7. The sign between 7 and 6 can be considered a plus (if it is a minus, we will change it and the next sign). Similarly, we can consider the sign between 6 and 5, then between 5 and 4, then between 4 and 3, and then between 3 and 2 as a plus.

If the sign between 2 and 1 is a plus, then Vanya has achieved the required: $7+6+5+4+3+2+1=$ 21 is divisible by 7.

If the sign between 2 and 1 is a minus, then Vanya can change the signs before 4 and 3, and then before 3 and 2. Then Vanya has again achieved the required: $7+6+5-4+3-2-1=14$ is divisible by 7.

## Criteria

The largest suitable criterion is used:

7 p. Any complete solution to the problem.

5 6. A correct algorithm of Vanya's actions is provided, but why it works is not obvious and not justified.

0 p. There is only the answer "cannot."