## Task 3 - 210723

Determine all pairs $(a ; b)$ of natural numbers $a$ and $b$ with $0<a<b$, whose greatest common divisor is 15 and whose product is 7875!
I. If a pair $(a ; b)$ of natural numbers $a$ and $b$ satisfies the conditions of the problem, then:

(1) There exist natural numbers $m, n$ such that $a=15 m, b=15 n$.

(2) Because $0<a<b$, it follows that $0<m<n$,

(3) because $a b=7875$, it follows that $15 m \cdot 15 n=7875$, thus $225 m n=7875, m n=35$. (3)

Since 35 has the prime factorization $35=5 \cdot 7$, for (2) and (3) there are only the possibilities that either $m=1, n=35$ or $m=5, n=7$. Therefore, from (1) it follows that only the pairs (15; 525), (75; 105) can satisfy the conditions of the problem.

II. These pairs satisfy these conditions; because $0<15<525,0<75<105$; due to the prime factorizations $15=3 \cdot 5,525=3 \cdot 5^2 \cdot 7,75=3 \cdot 5^2,105=3 \cdot 5 \cdot 7$, $3 \cdot 5=15$ is the gcd of 15 and 525 as well as the gcd of 75 and 105; finally, $15 \cdot 525=7875$ and $75 \cdot 105=7875$.

Therefore, exactly the pairs $(15 ; 525)$ and $(75 ; 105)$ satisfy the conditions of the problem.