Let $\sigma(n)$ be the sum of the divisors of $n$.
E.g. the divisors of $4$ are $1$, $2$ and $4$, so $\sigma(4)=7$.
The numbers $n$ not exceeding $20$ such that $7$ divides $\sigma(n)$ are: $4$, $12$, $13$ and $20$, the sum of these numbers being $49$.
Let $S(n, d)$ be the sum of the numbers $i$ not exceeding $n$ such that $d$ divides $\sigma(i)$.
So $S(20 , 7)=49$.
You are given: $S(10^6,2017)=150850429$ and $S(10^9, 2017)=249652238344557$.
Find $S(10^{11}, 2017)$.
Solution
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