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Math Help - sum of factors

  1. #1
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    sum of factors

    What is the sum of all factors of 45000 which are divisible by 10
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    That would be 10 times the sum of all the factors of 4500. The sum-of-factors function \sigma has the following property: if n=p_1^{k_1}\cdots p_r^{k_r} where the p_i are distinct primes, then

    \sigma(n)\ =\ \prod_{i\,=\,1}^r\frac{p_i^{k_i+1}-1}{p_i-1}

    Thus \sigma(4500)\,=\,\sigma(2^23^25^3)\,=\,\left(\frac  {2^3-1}{2-1}\right)\left(\frac{3^3-1}{3-1}\right)\left(\frac{5^4-1}{5-1}\right)\,=\,7\cdot13\cdot156\,=\,14196. So the answer to the problem is 141960.
    Last edited by TheAbstractionist; June 24th 2009 at 10:53 AM.
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  3. #3
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    can u give an easy method
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    I don’t see anything wrong with the proof, particularly the formula

    \sigma(n)\ =\ \prod_{i\,=\,1}^r\frac{p_i^{k_i+1}-1}{p_i-1}

    Shall we prove this formula? Then perhaps you’ll feel less afraid of using it.

    First, we prove that if \gcd(m,n)=1 then \sigma(mn)=\sigma(m)\sigma(n). Note that a divisor d of mn can be written as d=ef where e\mid m and f\mid n and \gcd(e,f)=1. Hence

    \sigma(mn)\,=\,\sum_{d\,\mid\,mn}d

    =\,\sum_{e\,\mid\,m\,f\,\mid\,n}ef

    =\,\sum_{e\,\mid\,m}e\,\sum_{f\,\mid\,n}f

    =\,\sigma(m)\sigma(n)

    Next, not that if p is a prime and k a positive integer, \sigma(p^k)=1+p+p^2+\cdots+p^k=\frac{p^{k+1}-1}{p-1}.

    Combining the two results established above gives the required formula for the sum-of-factors function \sigma.
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