The GCD of $2$ numbers is $6$ and the product of the two numbers is $4320$. How many pairs of numbers exist, which satisfy the above condition.


I took GCD $\times$ LCM=Product of two numbers

6 . LCM=4320

From which I get LCM=720.

I am not reaching anywhere to the solution.

Can anyone guide me how to solve the problem?



You are right so far. If the two numbers are $a$ and $b$, we have $(a,b) = 6$, so

$(6m, 6n) = 6$ where $(m, n) = 1$. Now the remaining $720$ has to be distributed among $m$ and $n$.

$720 = 2^4 \cdot 3^2 \cdot 5$. We cannot have any factor of the same prime distributed, i.e. all the $2$'s must go either with $m$ or with $n$. Since there are $3$ distinct prime factors, there are $8$ possibilities of distributing them. Because we are interested in pairs, we eliminate symmetry by dividing $8$ by $2$ and get $\color{blue}{4}$ possibilities as the final answer.

  • $\begingroup$ For the sake of symmetry?How symmetry exist? $\endgroup$ – Jack Nov 1 '15 at 4:24
  • 1
    $\begingroup$ The pair (a,b) is same as the pair (b,a) when we are just picking out pairs of numbers and order does not matter.. So I am eliminating the double counting which arises in my approach. Hope this helps $\endgroup$ – Shailesh Nov 1 '15 at 4:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.