How to find product of factors of a number if the factors are perfect square?

Let M be the set of all the distinct factors of the number N = 6^5 * 5^2 * 10, which are perfect squares. Find the product of the elements contained in the set M.

N = 2^6 * 3^5 * 5^3

Even power of 2 : 0,2,4,6 = 4

Even power of 3 : 0,2,4 = 3

Even power of 5 : 0,2 = 2

No of factors of number N that are perfect square = 4* 3 * 2 = 24

How to find product of elements in M???

• hint: the exponents of the primes in the prime factorization of a perfect square must be even. – symplectomorphic Jan 8 '16 at 14:09
• Yaa i know the steps till even power...check the edit – Aishwarya R Jan 8 '16 at 14:13

The elements $M$ come from choosing one power of each of $2,3,5,$ and multiplying them. The choices for the power of $2$ are, as you say, $1,4,16,64$ You can use the distributive property to add these up and multiply by the sum of the choices of $3$ and the sum of choices of $5$. In this case the sums do not have many terms, so you can just do it by hand. If there was a large list for one or more powers you could notice they are geometric series.