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Thread: Factorial problem help!

  1. #1
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    Factorial problem help!

    Can't see an easy way into this...

    Take 1!x 2!x3! x......100!

    Which factorial can you remove ( say divide by) so the result is a square number?

    Any thoughts?
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  2. #2
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    Re: Factorial problem help!

    You are referring to removing a single factorial? In order that something be a square, every prime divisor must occur an even number of times. How many multiples of 2 occur in each of those factorials? How many factors of 3? ....
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  3. #3
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    Re: Factorial problem help!

    Yes a single factorial. So if it was 1!x2!x3!x4! , removing the 2! would lead to a square number because it would become 1!x3!x3!x2x2 which is square.
    This is the way I have been thinking about it but got stuck .
    Last edited by rodders; Nov 8th 2016 at 02:27 PM.
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  4. #4
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    Re: Factorial problem help!

    Quote Originally Posted by rodders View Post
    Yes a single factorial. So if it was 1!x2!x3!x4! , removing the 2! would lead to a square number because it would become 1!x3!x3!x2x2 which is square.
    This is the way I have been thinking about it but got stuck .
    Why remove it?

    1! * 2! * 2! * 3! * 3! = 144

    Btw, * is now the standard multiplication symbol.
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  5. #5
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    Re: Factorial problem help!

    100!99! = 100 (99!)^2

    98!97! = 98 (97!)^2

    etc.

    Multiply the above to get

    1!2!3!\text{...}\text{..}100! = 2^{50}50!m^2

    (m an integer)

    so

    \frac{1!2!3!\text{...}\text{..}100! }{50!} is a perfect square
    Thanks from rodders, Plato and topsquark
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  6. #6
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    Re: Factorial problem help!

    Wow... cheers Thanks!
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