Number of perfect square divisors Calculate the number of perfect squares which are factors of 1!2!3!...100!
I tried to produce the prime factorisation of the product and then combine its factors but messed up.
For example, starting from the right, we definitely have 99^2, 98^2 and so on but what about the rest of their factors?
 A: Let $p_i$ be the various prime factors of the product, and $q_i=p_i^2$ (we have $p_i=2,3, \dots 97)$
We can write the product as $$\Pi q_i^{r_i}\Pi p_i^{\epsilon_i}$$ with $\epsilon_i\in \{0,1\}$ so that the product of the $p_i^{\epsilon_i}$ is square free.
Then any product of the $q_i$ (including the empty product) is a square by construction, and the number of such products is the product of terms of the form $r_i+1$. You simply need to use what you know of the prime factorisation here to determine the $r_i$.

So what you get is a prime factorisation of form $2^{2r_1+\epsilon_1}\cdot 3^{2r_2+\epsilon_2}\cdot \dots 97^{2r_{25}+\epsilon_{25}}$ where the epsilons are $1$ or $0$. There are $r_1+1$ ways of choosing an even power of $2$, $r_2+1$ ways of choosing an even power of $3$ etc and when you multiply even powers of primes together you get a square.
Now when you create the product $(r_1+1)(r_2+1)\dots (r_{25}+1)$ you are counting the ways of combining the different primes raised to even powers - i.e. the number of possible square factors.
