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- Feb 14, 2012

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

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- Thread starter
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- #1

- Feb 14, 2012

- 3,894

-----

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

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

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- #2

- Feb 14, 2012

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1. castor28

2. lfdahl

3. kaliprasad

4. Oxide

Solution from Oxide :

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 :

$$

\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 $16\times 7\times 3\times 2 = \bf 672$.

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