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Math Help - Perfect Square of a Product of Factorials Proof

  1. #1
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    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?
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  2. #2
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    Hello, ride12!

    given: . 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 N.

    \begin{array}{cc}\text{Factor} & \text{Frequency} \\ \hline<br />
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 2\cdot4\cdot6\cdot8\cdots 100 \;=\;2^{50}(1\cdot2\cdot3\cdot4\cdots50)

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


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

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  3. #3
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    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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