9.1. Given natural numbers $M$ and $N$, both greater than ten, consisting of the same number of digits, and such that $M = 3N$. To obtain the number $M$, one of the digits of $N$ must be increased by 2, and each of the other digits must be increased by an odd digit. What digit could the number $N$ end with? Find all possible answers.

(n. Agakhanov)
Answer. The digit 6.

Solution. By the condition, $M=3 N$, so the number $A=M-$ $-N=2 N$ is even. However, by the condition, the number $A$ is composed of odd digits and the digit 2. Therefore, $A$ ends in 2. Thus, the number $N$, which is half of $A$, ends in either 1 or 6.

We will show that $N$ cannot end in 1. If $N$ ends in 1, then when it is doubled, there is no carry from the last digit to the second-to-last digit. This means that the second-to-last digit of the number $A=2 N$ will be even, but it must be odd. This is a contradiction.

Remark. Pairs of numbers $N$ and $M$ that satisfy the condition do exist, for example, $N=16, M=48$. Moreover, there are infinitely many such pairs. All suitable numbers $N$ can be described as follows: the first digit is 1 or 2, followed by several (possibly zero) digits, each of which is 5 or 6, and the last digit is 6.

Comment. The correct answer and an example of a number $N$ ending in the digit 6 - 1 point.

It is established that the last digit of the number $M$ is 2 more than the last digit of the number $N$ - 1 point.

It is shown that the last digit of the number $N$ can only be 1 or 6 - 2 points.

Points for different advancements are cumulative.

Note that the problem does not require providing an example of such a number. It is sufficient to prove that no digit other than 6 can be the last digit.