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Doing more practice for my final and I need some help with the following:

Evaluate: $$\sum_{d\mid2020}{\sigma(d)}$$

where $\sigma(n)$ is the sum of all divisors of n.

The hint given specifically states that this isn't asking for the sum of all the divisors of 2020!.

I found $${\sigma(2020)} = 4284$$ but I need some help progressing from here. Thanks!

Shanker
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    Use the multiplicativeness of $f(n)=\sum_{d | n} \sigma(d)$ that is $f(n) = \prod_{p^k | n}f(p^k)= \prod_{p^k | n} \sum_{m=0}^k \sigma(p^m)=\prod_{p^k | n} \sum_{m=0}^k \frac{p^{m+1}-1}{p-1}$. Then evaluate this with $n = 2020 = 101 . 2^2 . 5$ – reuns Jun 02 '19 at 03:44
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    It should be $\sigma(n)$ is the sum of all the divisors of $n$ – Ross Millikan Jun 02 '19 at 04:17
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    I've been answering this kind of problems in the past days, here https://math.stackexchange.com/questions/3246978/for-gn-sum-dnfd-find-g5000/3247014#3247014 also here https://math.stackexchange.com/questions/3245975/liouville-function-and-perfect-square-2?noredirect=1#comment6675730_3245975 – Julian Mejia Jun 02 '19 at 05:51

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The convolution of two multiplicative functions is multiplicative. Since $\sigma=id\ast 1$, we have $\sigma$ is multiplicative. Even more, we have $f(n)=\sum_{d|n}\sigma(d)$ is multiplicative($f=\sigma\ast 1$, the convolution of two multiplicative functions).

Multiplicative means that $f(mn)=f(m)f(n)$ when $\gcd(m,n)=1$. In particular a multiplicative function is determined by the values of it at prime powers.

In particular, we have $2020=(2^2)(5)(101)$. Hence, $$f(2020)=f(2^2)f(5)f(101)=(\sigma(1)+\sigma(2)+\sigma(2^2))(\sigma(1)+\sigma(5))(\sigma(1)+\sigma(101))=(1+3+7)(1+6)(1+102)=7931$$

Julian Mejia
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