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Math Help - number of factorisations

  1. #1
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    number of factorisations

    How many different non-trivial factorisations does

    3 5 7 11 13 17 19 23 29 31 = 100280245065.

    have?

    Since there are 10 primes, i thought the answer would be

    sum_(i=0)^n {10 \choose i} - 1 = 2^10 - 1

    but i am a little bit unsure if i am not counting anything double.
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  2. #2
    Super Member Bacterius's Avatar
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    Try with 3 \times 5 \times 7 and 3 \times 5 \times 7 \times 11 (do it by hand, boring but useful), look for a pattern, check it with 5 prime factors, write it in maths, then apply it on your 10 prime factors
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  3. #3
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    I was wondering if the following is a correct argument:

    Let n be the number as before and consider the set A:={3,5,7,11,13,17,19,23,29,31}. The product of the elements of each subset B of this set give a different divisor of the given number. However, the factorisation obtained by diving b by the product of the elements in some subset B, is the same as the factorisation obtained by diving n by the product of the elements in A\B.

    The total number of subsets of A is equal to the number of elements in P(A)=power set of A. However, this includes the empty set and A itself, which yield a trivial factorisation. Thus the total number of non-trivial factorisations is 1/2*(2^(10)-2)=2^9-1.
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  4. #4
    Super Member Bacterius's Avatar
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    It seems to work with n = 3, because we have manually 3 arrangements, and with your formula : 2^{n - 1} - 1 = 2^2 - 1 = 4 - 1 = 3
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