# 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).

Jul 21, 2016

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:
${90}^{9} = {\left(5 \cdot 3 \cdot 3 \cdot 2\right)}^{9}$
$= {\left(5 \cdot {3}^{2} \cdot 2\right)}^{9}$
$= {5}^{9} \cdot {3}^{18} \cdot {2}^{9}$

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:
${5}^{9} \cdot {3}^{18} \cdot {2}^{9}$

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, ${\left({2}^{2} \cdot {5}^{2}\right)}^{2}$ is a square of an integer, as is ${\left({3}^{8} \cdot {2}^{4}\right)}^{2}$; 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 $5 \cdot 10 \cdot 5$, 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.