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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.

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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.

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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}

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