7. (50 points) Two multiplication examples of two natural numbers were written on the board, and in one of the examples, the numbers were the same. The troublemaker Petya erased both the original numbers and almost all the digits of the multiplication results, leaving only the last three digits: ... 689 and ... 759. Can the excellent student Misha determine which of these three digits belong to the number that resulted from the multiplication of the same numbers?
Answer: It could be 689.

First solution. Statement: at least one of the two last digits of the square of a natural number is necessarily even. Then the second number is eliminated. Example for the first number: $133 \cdot 133=17689$.

Proof of the statement: the statement is true for even numbers, let's consider the squares of odd numbers. Let the second-to-last digit of the original number be $n$, the last digit be $m$, where $m \in\{1,3,5,7,9\}$, and all other digits of the number be $k$. Then

$$
\left[\begin{array}{l}
(100 k+10 n+1)^{2}=100 \cdot(\ldots)+10 \cdot 2 n+1 \\
(100 k+10 n+3)^{2}=100 \cdot(\ldots)+10 \cdot 6 n+9 \\
(100 k+10 n+5)^{2}=100 \cdot(\ldots)+10 \cdot 2+5 \\
(100 k+10 n+7)^{2}=100 \cdot(\ldots)+10 \cdot(4 n+4)+9 \\
(100 k+10 n+9)^{2}=100 \cdot(\ldots)+10 \cdot(8 n+8)+1
\end{array}\right.
$$

In all cases, the second-to-last digit of the square is even.

Second solution. We will base our reasoning on the following statement: the square of a natural number is either divisible by 4 or gives a remainder of 1 when divided by 8. Indeed, if the original number is even, then its square is divisible by 4, and if the original number is odd, then it can be represented as $8 q+r$, where $r \in\{1,3,5,7\}$. Then $(8 q+$ $+r)^{2}=8 \cdot(\ldots)+1$. Having the last three digits of the number in front of us, we can check the divisibility rule by 4 or find the remainder when the number is divided by 8.