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 with $n$ . (For example, $\phi(6)=2$ and $\phi(10)=4$ .) Let $S=\sum_{d|2008}\phi(d)$ , where $d$ 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 all the divisors of an integer n are { ${d_1}$ , ${d_2}$ , ${d_3}$ , ... ${d_k}$ }, then n = $\phi({d_1})$ + $\phi({d_2})$ + .... + $\phi({d_k})$ . This provides a shorter approach in this problem because it asks about the sum of the number of numbers co prime to 2008's divisors, as mentioned earlier. This is 2008 itself.

Proof

Let $n$ = ${p_1}^{\alpha_1}$ ${p_2}^{\alpha_2}$ .... ${p_k}^{\alpha_k}$ where ${p_1}, {p_2}, ... {p_k}$ are prime numbers

and

{ ${d_1}$ , ${d_2}$ , .... ${d_m}$ } is an exhaustive, mutually exclusive list of $n$ 's factors

$\phi({d_1})$ + $\phi({d_2})$ + ... + $\phi({d_m})$

= ${p_1}(1-\frac{1}{p_1})$ + ${p_2}(1-\frac{1}{p_2})$ + ${p_1} {p_2}(1-\frac{1}{p_1})(1-\frac{1}{p_2})$ ...

The above term represents the sum of all the values of the the totient function applied of all of n's divisors, like the problem is asking.

= $(1 - \frac{1}{p_1})$ $(1 - \frac{1}{p_2})$ $({p_1} + {p_1}^{2} + ... + {p_1}^{\alpha_1})$ $({p_2} + {p_2}^{2} + ... + {p_2}^{\alpha_2})$ + ....

= $(1 + (1 - \frac{1}{p_1})$ $({p_1} + {p_1}^{2} + ... + {p_1}^{\alpha_1})$ ) $(1 + (1 - \frac{1}{p_2})$ $({p_2} + {p_2}^{2} + ... + {p_2}^{\alpha_2})$ ) ... $(1 + (1 - \frac{1}{p_k})$ $({p_k} + {p_k}^{2} + ... + {p_k}^{\alpha_k})$

= $(1 + {p_1} + {p_1}^{2} + ... + {p_1}^{\alpha_1} - (1 + {p_1} + {p_1}^{2} + ... + {p_1}^{\alpha_1 - 1}))$ x

$(1 + {p_2} + {p_2}^{2} + ... + {p_2}^{\alpha_2} - (1 + {p_2} + {p_2}^{2} + ... + {p_2}^{\alpha_2 - 1}))$ x ...

$(1 + {p_k} + {p_k}^{2} + ... + {p_k}^{\alpha_k} - (1 + {p_k} + {p_k}^{2} + ... + {p_k}^{\alpha_k - 1}))$

= ${p_1}^{\alpha_1}$ ${p_2}^{\alpha_2}$ ... ${p_k}^{\alpha_k}$

= $n$

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

Contents

Problem

Solution

Extra Note

Proof

See Also