## Task 2 - 050612

A two-digit natural number is to be determined based on the following conditions:

Its cross sum is 10. If you swap its digits and add 1 to the resulting number, you get double the original number.
If the number of tens is denoted by $a$ and the number of units by $b$, then the first number is $(10 a+b)$ and the second $(10 b+a)$. According to the problem, we have:

$$
\begin{aligned}
a+b & =10 \\
2(10 a+b) & =(10 b+a)+1
\end{aligned}
$$

The left number in (2) is even. If $a$ were even, the right number would be odd. Therefore, $a$ must be odd, and thus (because $a+b=10$) $b$ must also be odd.

The right side of (2) is at most equal to $10 b+10$. If $a$ were greater than or equal to $b$, then the left side would be at least equal to $2(10 b+b)=22 b=10 b+12 b$.

This is certainly greater than $10 b+10$ (because $b \geq 1$), so $b$ must be greater than $a$.

Thus, the only possible pairs of numbers are $a=1 ; b=9$ and $a=3 ; b=7$. By trying, one immediately finds that $a=3 ; b=7$ is the only pair that satisfies the conditions of the problem. Therefore, the number sought is 37.