Perfect Squares as Divisors of $1!\cdot 2! \cdot 3! \cdots 9!$

[anemone]

Here is this week's POTW:

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How many perfect squares are divisors of the product $1!\cdot 2! \cdot 3! \cdots 9!$?

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[anemone]

Congratulations to the following members for their correct solution!

1. castor28
2. lfdahl
3. kaliprasad
4. Oxide

Solution from Oxide:

Spoiler

By expanding out the product \(\displaystyle n=1! 2! \cdots 9!\), we get \(\displaystyle 2^8 3^7 4^6 5^5 6^4 7^3 8^2 9^1\), which can be factored into primes as \(\displaystyle 2^{30} 3^{13} 5^5 7^3\). Since a perfect square that divides \(\displaystyle n\) must be of the form \(\displaystyle 2^a 3^b 5^c 7^d\) where \(\displaystyle a,b,c,d\) are even, we can choose \(\displaystyle a,b,c,d\) from the following sets respectively:
\(\displaystyle \{0,2,\dots, 30\}\)
\(\displaystyle \{0,2,\dots,12\}\)
\(\displaystyle \{0,2,4\}\)
\(\displaystyle \{0,2\}\)
which gives us \(\displaystyle 16 \cdot 7 \cdot 3 \cdot 2 = 672\) perfect squares that divide \(\displaystyle n\).

Alternate solution from castor28:

Spoiler

We factorize $n!=2^\alpha3^\beta5^\gamma7^\delta$ for $2\le n\le9$ as follows (note that each line is easily computed from the previous line):
$$
\begin{array}{c|r|r|r|r}
n!&\alpha&\beta&\gamma&\delta\\
\hline
2!&1&0&0&0\\
3!&1&1&0&0\\
4!&3&1&0&0\\
5!&3&1&1&0\\
6!&4&2&1&0\\
7!&4&2&1&1\\
8!&7&2&1&1\\
9!&7&4&1&1\\
\hline
\prod{n!}&30&13&5&3
\end{array}
$$
The square divisors of the product are of the form $2^x 3^y 5^z7^w$, with $x$, $y$, $z$, $w$ even and $(x,y,z,w)\le (30,13,5,3)$.
Since the number of even integers between $0$ and $k$ is $\left\lfloor\dfrac{k}{2}\right\rfloor+1$, the number of square divisors of the product is $16\times 7\times 3\times 2 = \bf 672$.