## Problem 1.

Determine all natural numbers $a$ and $b$, where $a < b$, that are two-digit numbers, knowing that the greatest common divisor (gcd) of $a$ and $b$ is a prime number, 20 times smaller than the least common multiple (lcm) of $a$ and $b$.
Solution. Let $p=(a, b)$, a prime number. Then $a=p a_{1}, b=p b_{1}$ with $\left(a_{1}, b_{1}\right)=1, a_{1}<b_{1}$ (since $a<b$ by hypothesis). From the relation $a \cdot b=[a, b] \cdot(a, b)$ and the problem statement, we have: $20(a, b)=[a, b]$, i.e., $20 p=\frac{p a_{1} \cdot p b_{1}}{p}$, from which $a_{1} b_{1}=20$. Since $\left(a_{1}, b_{1}\right)=1, a_{1}<b_{1} \Rightarrow a_{1}=1, b_{1}=20$ or $a_{1}=4, b_{1}=5$.

I. Case $a_{1}=1, b_{1}=20$ does not work.

II. Case $a_{1}=4, b_{1}=5$. From $(a, b)=p$ prime $\Rightarrow p \geq 2$. But $a$ and $b$ are two-digit numbers $\Rightarrow p<20$, i.e., $p \in\{3,5,7,11,13,17,19\}$. We have the solutions:

$a=12, b=15 ; a=20, b=25 ; a=28, b=35 ; a=44, b=55 ; a=52, b=65 ;$ $a=68, b=85 ; a=76, b=95$.