Number Theory problem 1 by Dhaval Furia

$2^43^510^4=(4^49^225^2)3$ . So, in all, there are $(4+1)(2+1)(2+1)$ or $45$ such square factors, of which the number of such factors greater than $1$ is $\boxed {44}$ .

Prime factorization of 2^4 x 3^5 x 10^4 is 2^8 x 3^5 x 5^4

So in the prime factorization of a perfect square factor of this number, the power of 2 can be anything from 0, 2, 4, 6, 8 (5 choices)

Similarly, that of 3 can be anything from 0, 2, 4 (3 choices)

And that of 5 can be anything from 0, 2, 4 (3 choices)

Hence, (by Multiplication Counting Principle), there are 5x3x3 = 45 possible perfect square factors

But this also includes 1 (when powers of 2, 3 and 5 are all 0)

Since we want the count of perfect squares greater than 1, it will be 45 - 1 = 44