11.1. A natural number greater than 1000000 gives the same remainder when divided by 40 and by 625. What digit can stand in the thousands place of this number? (N. Agakhanov, K. Sukhov)
Answer: 0 or 5.

Solution. Let $n$ be the given number, and $t$ be its remainder when divided by 40 and 625. Then the number $n-t$ is divisible by 40 and 625, which means it is divisible by $\operatorname{LCM}(40; 625)=5000$. Therefore, the difference $n-t$ ends in either 5000 or 0000. The remainder $t<40$. Thus, the digit in the thousands place can be 0 or 5. Both situations are possible, for example, the numbers 20210000 and 20215000 (both of these numbers have a remainder of 0 when divided by 40 and 625).

Comment. Correct answer without explanation - 0 points.

Only one example with one of the two digits - 0 points.

Only examples with the digits 0 and $5-1$ point.

Proven only that the digit in the thousands place can be 0 or $5-5$ points.