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Re: What is the sum of ALL the factors of 49?
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16 Nov 2016, 06:06

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When an integer n can be expressed as \(a^p*b^q*c^r\), where a,b,c are prime factors of n and p,q,r are their powers, the sum of the factors of n can be expressed as \(\frac{(a^p+1-1)*(b^q+1-1)*(c^r+1-1)}{(a-1)*(b-1)*(c-1)}\)

49=\(7^2\) Applying the formula, \(\frac{(7^2+1-1)}{(7-1)=(7^3-1)/6}\) = 57 So, answer E.
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