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

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.