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    • Thread Starter

    Find its prime factorisation, add 1 to all the powers in it and multiply them together (this includes 1 as a factor) e.g.

    2500 = 5^4 x 2^2 gives 5 x 3 = 15 factors (including 1 and itself)

    The product gives the amount of combinations of products of factors i.e. the amount of numbers that 2500 is divisible by because they are all of the form 5^m x 2^n for m=0,1,2,3,4 n=0,1,2

    I think it does work all the time but I'm just checking. Certainly a lot easier than finding all the factors individually so might be a good alternative to use in future.
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    (Original post by oh_1993)
    Yes. If you express it as p_1^{a_2}p_2^{a_1}...p_n^{a_n} where all the p_i are distinct.

    I assume you're not interested in negative factors, just +ve ones.
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Updated: January 2, 2013
“Yanny” or “Laurel”
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