# 2008 iTest Problems/Problem 39

## Problem

Let $\phi(n)$ denote $\textit{Euler's phi function}$, the number of integers $1\leq i\leq n$ that are relatively prime to $n$. (For example, $\phi(6)=2$ and $\phi(10)=4$.) Let $S=\sum_{d|2008}\phi(d)$, in which $d$ ranges through all positive divisors of $2008$, including $1$ and $2008$. Find the remainder when $S$ is divided by $1000$.

## Solution

The divisors of $2008$ are $1$, $2$, $4$, $8$, $251$, $502$, $1004$, and $2008$, so the problem requires the sum of the number of numbers relatively prime to each of the eight numbers.

Using the formula $\phi(n) = n \cdot (1 - \tfrac{1}{p_1})(1 - \tfrac{1}{p_2}) \cdots (1 - \tfrac{1}{p_n})$ (or by manually counting the numbers relatively prime to each number), $1$ number is relatively prime to $1$, $1$ number is relatively prime to $2$, $2$ numbers are relatively prime to $4$, $4$ numbers are relatively prime to $8$, $250$ numbers are relatively prime to $251$, $250$ numbers are relatively prime to $502$, $500$ numbers are relatively prime to $1004$, and $1000$ numbers are relatively prime to $2008$. That means $S = 2008$, and the remainder when $S$ is divided by $1000$ is $\boxed{8}$.