Given GCD of two numbers is 42 and their product is 15876. How many possible sets of members can be found?

Question

Given GCD of two numbers is 42 and their product is 15876. How many possible sets of numbers can be found?

I have no idea. I can only evaluate the lcm. Don't know how to get the answers.

Answer 1

You have two numbers that are divisible by $42$. We can write these as $42a$ and $42b$ for $a, b$ coprime (can you see why?). Their product is $15876$, so we have $$ 42^{2}ab = 15876 \iff ab = 9. $$ It it now enough to determine which $a, b$ satisfy $ab = 9$, where $a$ and $b$ are coprime.

Answer 2

Use $\gcd(x,y) \times \operatorname{lcm}(x,y) = xy$

\begin{array}{rrr} xy &= &15876 \\ \operatorname{lcm}(x,y) &= &378 \\ \hline \gcd(x,y) &= &42 \end{array}

Assume $x < y$. Use $\gcd(p^a, p^b) = p^{min(a,b)}$ and $\operatorname{lcm}(p^a, p^b) = p^{max(a,b)}$ when $p$ is a prime number.

\begin{array}{rcr|ccc} \gcd(x,y) &= &42 & 2^1 & 3^1 & 7^1 \\ \operatorname{lcm}(x,y) &= &378 & 2^1 & 3^3 & 7^1 \\ xy &= &15876 & 2^2 & 3^4 & 7^2 \\ \hline x &= &42 & 2^1 & 3^1 & 7^1 \\ y &= &378 & 2^1 & 3^3 & 7^1 \\ \end{array}