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  1. #1
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    divisor

    For every positive even integer n, the function H(n) is defined to be the product of all even integers from 2 to n. What is the lowest possible prime factor of H(100) =1.
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    Quote Originally Posted by chibuike1 View Post
    For every positive even integer n, the function H(n) is defined to be the product of all even integers from 2 to n. What is the lowest possible prime factor of H(100) =1.
    Do you mean what is the lowest possible prime factor of H(100) - 1?

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chibuike1 View Post
    For every positive even integer n, the function H(n) is defined to be the product of all even integers from 2 to n. What is the lowest possible prime factor of H(100) =1.
    H(100) = 2 \cdot 4 \cdot 6 \cdot ~ ... ~ \cdot 98 \cdot 100

    = 2^{50} \cdot (1 \cdot 2 \cdot 3 \cdot ~ ... ~ \cdot 49 \cdot 50 )

    So H(100) - 1 cannot be divisible by any prime less than 53, the first prime which does not appear on this list. I'm not a number theorist so there may be a way to prove that this is not divisible by a prime higher than this.

    -Dan
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  4. #4
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    Math solution reqd

    Sorry the right question is:

    For every positive even integer n, the function H(n) is defined to be the product of all even integers from 2 to n. What is the lowest possible prime factor of H(100) + 1.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chibuike1 View Post
    Sorry the right question is:

    For every positive even integer n, the function H(n) is defined to be the product of all even integers from 2 to n. What is the lowest possible prime factor of H(100) + 1.
    From my knowledge base, the answer would be the same as the one I gave for H(100) - 1.

    -Dan
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