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Math Help - 10^2006|(n^2-n) for 0<n<10^2006

  1. #1
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    10^2006|(n^2-n) for 0<n<10^2006



    is it no such number existed?
    cos the minimium n is 10^2006?

    if so, how to prove it?
    please help, thanks.
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  2. #2
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    10^x will be a product of 2s and 5s of equal proportion.

    n^2 - n = n(n-1)

    So we're looking for consecutive integers, one of which is odd.

    The odd number and even number must each divide 10^x.
    The odd factor must be a product of only 5s. It's "one's digit" will be 5. The even number must be a product of 2s and then may also be a product of 2*5s. If the even number contains any 2*5 factors, it's "one's digit" will be 0, which is not adjacent to an odd number with a "one's digit" of 5. So the only hope rests with an equal distribution of 2s to the even number and 5s to the odd number. As there is no number x wherein 2^x and 5^x have a difference of one, there is no 10^x that is divisible by n^2 - n.
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