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Just the Evens

Find the sum of even positive divisors of 100000

From the arithmetic argument established previously, we know that the sum of divisors function is defined as

    \[\sigma(n) = \prod_{p|n, p \mbox{\tiny{ prime}}} \frac{p^{e_p+1}-1}{p-1}\]

Because 100000 = 2^5 5^5, then it must be the case that \sigma(n) = \frac{2^6-1}{2-1} \frac{5^6-1}{5-1}; however, because we can only look at the even divisors, we can only consider exponents of 2 that are greater than or equal to 1. That is, using the same argument used in the previous article

    \begin{align*} \sigma_{even}(10^5) &= (2^1 + 2^2 + 2^3 + 2^4 + 2^5) (5^0 + 5^1 + 5^2 + 5^3 + 5^4 + 5^5) \\ &= \sigma(10^5) - \sigma(5^5) \end{align*}

Therefore, \sigma_{even}(10^5) = (2^6-2) \frac{5^6-1}{4}. \blacksquare

Posted By Lee

Woosh