Question

The number `90^9` has `1900` different positive integral divisors. How many of these are squares of integers?

The answer is 250 but I'm not sure how to get there. (Number theory isn't one of my strong suits).

Answer 1

Wow - I get to answer my own question.

Explanation:

It turns out that the approach is a combination of combinatorics and number theory. We begin by factoring `90^9` into its prime factors:
`90^9=(5*3*3*2)^9`
`=(5*3^2*2)^9`
`=5^9*3^18*2^9`

The trick here is to figure out how to find squares of integers, which is relatively simple. Squares of integers can be generated in a variety of ways from this factorization:
`5^9*3^18*2^9`

We can see that `5^0`, for example, is a square of an integer and a divisor of `90^9`; likewise, `5^2`, `5^4`,`5^6`, and `5^8` all meet these conditions as well. Therefore, we have 5 possible ways to configure a divisor of `90^9` that is a square of an integer, using 5s alone.

The same reasoning applies to `3^18` and `2^9`. Every even power of these prime factors - 0, 2, 4, 6, 8, 10, 12, 14, 16, 18 (10 total) for 3 and 0, 2, 4, 6, 8 (5 total) for 2 - is a perfect square who is a divisor of `90^9`. Furthermore, any combination of these prime divisors who have even powers also satisfies the conditions. For instance, `(2^2*5^2)^2` is a square of an integer, as is `(3^8*2^4)^2`; and both, being made up of divisors of `90^9`, are also divisors of `90^9`.

Thus the desired number of squares of integers that are divisors of `90^9` is given by `5*10*5`, which is the multiplication of the possible choices for each prime factor (5 for 5, 10 for 3, and 5 for 2). This is equal to `250`, which is the correct answer.