# How many pairs of numbers exist,which satisfy the following conditions?

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.

MyApproach

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?

## 1 Answer

Outline

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.

• For the sake of symmetry?How symmetry exist? – Jack Nov 1 '15 at 4:24
• 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 – Shailesh Nov 1 '15 at 4:27