$:$ Folkiore

Given two two-digit numbers $-X$ and $Y$. It is known that $X$ is twice as large as $Y$, one digit of the number $Y$ is equal to the sum, and the other digit is equal to the difference of the digits of the number $X$.

Find these numbers.
Let $a$ be the first digit of the number $X$, and $b$ be the second. The first digit of the number $Y$ is less than $a$, so it is equal to $|a-b|$, and the second digit is $a+b$. We can reason in different ways.

First method 1. Let's consider 2 cases.

1) $10 a+b=20(a-b)+2(a+b)$. Then $12 a=19 b$, which is impossible: $a$ does not divide by 19.
2) $10 a+b=20(b-a)+2(a+b)$. Then $28 a=21 b$, that is, $4 a=3 b$, and it remains to check the numbers 68 and 34.

Second method. By comparing the last digits of the numbers $X$ and $Y$, we notice that $2(a+b) \equiv b(\bmod 10)$. Therefore, $2 a+b$ is divisible by 10. Since $a+b<10$, then

$2 a+b=10$. It remains to check the numbers $18,26,34,42,50$.

## Answer

34 and 17.