Sum of perfect square divisors

Question

Please again help me with another question

Sum of perfect square divisors of $2^3 \cdot 21^2\cdot 91^1$ is $A$, then number of divisors of $A$ is?

How to do it and what is its answer? Thnx

Answer 1

First, we factor the number $N=2^3\cdot 21^2\cdot 91^1$, and see it is $N=2^3\cdot3^2\cdot7^3\cdot13$. A number is square if and only if all its prime factors occur an even amount of times. It is therefore sufficient to find the sum of square divisors of $M=2^2\cdot3^2\cdot7^2$, and it is not hard to see that this is $$(1+2^2)(1+3^2)(1+7^2)=5\cdot10\cdot50=2500$$

Thus, $A=2500$. Now as for the number of divisors of $A$, note that $A=2^2\cdot 5^4$. I assume you can solve the rest yourself now, after all, I don't want to spoil everything for you ;)