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The title says it all, there is a product, as shown above, one of the factorials must be removed and the product will make a perfect square. Which one?

For example, you could remove $54!$?

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5 Answers 5

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Group them as $$\begin{align*}(1!2!)(3!4!)\cdots (99!100!) &= (1!^2\times 2)(3!^2 \times 4)\cdots (99!^2\times 100) \\&= (1!3!\cdots 99!)^2 \times 2\cdot 4\cdot 6\cdots 100 \\ & = (1!3!\cdots99!)^2\times 2^{50} \times 50! \end{align*} $$

so you can remove $50!$ and it's a square.

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Hint. Show that the number to remove is $50!$ (note that $50$ is half of $100$).

P.S. More generally, if $m$ is a positive integer then $$N:=\prod_{j=1}^{4m}(j!)=\prod_{k=1}^{2m}\left[(2k-1)!(2k)!\right]=\left(2^m\prod_{k=1}^{2m}(2k-1)!\right)^2\cdot (2m)!$$ which implies that $N/(2m)!$ is a perfect square.

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    $\begingroup$ As an aside, in the case where $2m$ is a square itself, you get two possible solutions. $\endgroup$
    – Zain Patel
    Commented Jun 23, 2017 at 12:44
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What you need to do is keep track of unbalanced occurrences of prime factors. Since $(2n)!=(2n-1)!(2n)$, every pair of these consecutive factors $((2n-1)!, \ (2n)!)$ contributes the same as a single factor $2n$ to the unbalance of prime factors. Since we can pair up all $100$ factorials into $50$ such pairs, the unbalance is the same as that of the product $2\times 4\times\cdots\times 98\times 100$, which product equals $2^{50}50!$. Since $2^{50}$ is a perfect square, you can remove all unbalance by removing the factor $50!$ from the product.

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Here is a tabular approach. Make up the table: enter image description here

Note that in even columns the degree is even (e.g. $3^{98}$). From odd columns we will take away by one number and collect them to get $2\cdot4\cdots100=2^{50}\cdot50!$. Hence $50!$ must be removed to make the product a square.

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First of all, you can write every $(n!)$ as $(n-1)!\cdot n$.

However, for now, just do this for all the odd values of $n$ and you get

$$(2!)(2!)\cdot 3 \cdot(4!)(4!)\cdot 5\cdots (98!)(98!)\cdot 99\cdot 100!$$

Now call $(2!)(4!)\cdots (98!)=C$ to simplify and your expression is equal to

$$C^2\cdot 3\cdot 5\cdot 7\cdots 99\cdot (100!)$$

Write out $100!$ and that's equal to

$$C^2\cdot 3\cdot 5\cdot 7\cdots 99\cdot 2\cdot 3\cdot 4\cdots 100$$

Rearrange order to get

$$C\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7 \cdot 7\cdots 99\cdot 2\cdot 4\cdot 6\cdots 100$$

Now call $3\cdot 5\cdot 7\cdots 99=D$ and that's equal to

$$C^2D^2 2\cdot 4\cdot 6\cdots 100 = C^2D^2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 2\cdot 4\cdots 2\cdot 50$$

Group all the twos together and you see that's equal to $$C^2D^22^{50}\cdot 50!$$

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