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 $\boxed{8}$ .

2008 iTest (Problems)
Preceded by: Problem 38	Followed by: Problem 40
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2008 iTest Problems/Problem 39

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