Competition problem — Number theory

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Number theory Difficulty 4.6 AIME Find the answer

Compute $\sum_{k=1}^{2009} k\left(\left\lfloor\frac{2009}{k}\right\rfloor-\left\lfloor\frac{2008}{k}\right\rfloor\right)$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The summand is equal to $k$ if $k$ divides 2009 and 0 otherwise. Thus the sum is equal to the sum of the divisors of 2009, or 2394.

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Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.