# Sum of Sum of Divisors

For all $$n \in \mathbb{N}$$, let $$\tau (n)$$ denote the sum of positive divisors of $$n$$ (including $$1$$ and itself). Find the last three digits of $$\displaystyle \sum \limits_{n=1}^{100} \tau (n)$$.

Details and assumptions

• As an explicit example, the positive divisors of $$4$$ are $$1, 2$$ and $$4$$, so $$\tau (4)= 4+2+1 = 7$$.

• This is not a computer science problem.

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