$2-$

Find the numbers equal to twice the sum of their digits.

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If the first digit of a two-digit number is $a$, and the second digit is $b$, then the number itself is $10a + b$.

## Solution

It is easy to notice that there are no single-digit numbers greater than zero with the required property. Let's try to find a solution among two-digit numbers. If the first digit of a two-digit number is $a$, and the second digit is $b$, then the number itself is $10a + b$. We have $10a + b = 2(a + b)$. From this, $8a = b$, so $a = 1, b = 8$.

We can show that there are no other solutions (idea: the smallest three-digit number is 100, and the largest sum of three digits is $9 + 9 + 9 = 27$). However, this was not required on the olympiad.

## Answer

$0 ; 18$.