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 in $N$ 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?
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.

If $N$ ends in 1, then when it is doubled, there is no carry from the last digit to the second last digit. This means the second last digit of the number $A=2 N$ would be even, but it must be odd. This is a contradiction.

## Answer

The digit is 6.