# Thread: Perfect Square of a Product of Factorials Proof

1. ## Perfect Square of a Product of Factorials Proof

Sorry about the double post if some of you are reading both sections but I was unsure where this could go, so I decided to ask in both.

Here is an interesting proof (in my opinion) maybe some of you will have some insight. I have managed to prove it but in a long and non efficient way. Here's the proof:

We have 1!*2!*3!*4!*5!*6!...*100!

Prove that you can eliminate one factor so that the remaining number is a perfect square.

What I have done is using the help of a computer prime factor each number 1! through 100!, then added their powers and from that I saw that by eliminating 50! the number 1!*2!*3!*4!*5!*6!...*100! will have its prime factorization all to an even power, resulting in a perfect square. Now I am sure there should be a more efficient way in doing this, any help?

2. Hello, ride12!

given: .$\displaystyle N \;=\;1!\cdot2!\cdot3!\cdot4!\cdot5!\cdot6! \cdots 100!$

Prove that you can eliminate one factor so that the remaining number is a perfect square.

Consider the frequency of each factor in the number $\displaystyle N.$

$\displaystyle \begin{array}{cc}\text{Factor} & \text{Frequency} \\ \hline 1 & 100 \\ 2 & 99 \\ 3 & 98 \\ 4 & 97 \\ \vdots & \vdots \\ 98 & 3 \\ 99 & 2 \\ 100 & 1 \end{array}$

The odd factors {1, 3, 5, ... , 99} all appear an even number of times.
. . They will comprise a square number.

The even factors {2, 4, 6, ..., 100} all appear an odd number of times.
. . If we eliminate one of each, the rest will comprise a square number.

But $\displaystyle 2\cdot4\cdot6\cdot8\cdots 100 \;=\;2^{50}(1\cdot2\cdot3\cdot4\cdots50)$

So we can eliminate just $\displaystyle 1\cdot2\cdot3\cdot4\cdots50 \:=\:\boxed{50!}$
. . and leave the $\displaystyle 2^{50}$, a square.

Then: .$\displaystyle \frac{N}{50!} \;=\;\left(1^{100}\cdot 3^{98}\cdot 5^{96} \cdot 7^{94} \cdots 99^2\right)\cdot 2^{50}$ . . . a square

3. Brilliant!
Thank you very much for the help. I was just considering the odd and even factors when I refreshed the forum page, but you pointed me in the right direction

Thank again, and continue the great work

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# Consider the product N = 1! × 2! × 3! × ... × 100!. Is it possible to remove one of the terms from this product and have the remaining product be a perfect square?

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