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2008 iTest Problems/Problem 39

Revision as of 14:43, 20 August 2023 by Tikachaudhuri (talk | contribs) (→‎Solution)

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 $\boxed{8}$.

Extra Note

If the divisors of n are ${d_1}$, ${d_2}$, ${d_3}$, ... ${d_k}$, $\phi(n)$ = $\phi({d_1})$ + $\phi({d_2})$ + .... + $\phi({d_k})$. This provides a shorter approach in this problem because it asks about

See Also

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