Difference between revisions of "2008 iTest Problems/Problem 39"

Revision as of 07:04, 21 August 2023

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 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 the same as 2008 itself, drastically simplifying the problem.

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.

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

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

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

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

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

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

= $n$

While this may seem like such a complex proof for this problem, numbers may have far more factors than 2008, and testing it out may be difficult.

@@ Line 16: / Line 16: @@
 ==Proof==
-Given that: <math>n</math> = <math>{p_1}^{\alpha_1}</math><math>{p_2}^{\alpha_2}</math>....<math>{p_k}^{\alpha_k}</math>
+Let <math>n</math> = <math>{p_1}^{\alpha_1}</math><math>{p_2}^{\alpha_2}</math>....<math>{p_k}^{\alpha_k}</math> where <math>{p_1}, {p_2}, ... {p_k}</math> are prime numbers
 and
-<math>{d_1}</math>, <math>{d_2}</math>, .... <math>{d_m}</math> are all of <math>n</math>'s divisors
+{<math>{d_1}</math>, <math>{d_2}</math>, .... <math>{d_m}</math>} is an exhaustive, mutually exclusive list of <math>n</math>'s factors
 <math>\phi({d_1})</math> + <math>\phi({d_2})</math> + ... + <math>\phi({d_m})</math>

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