2008 iTest Problems/Problem 39
Contents
Problem
Let denote , the number of integers that are relatively prime with . (For example, and .) Let , where ranges through all positive divisors of , including and . Find the remainder when is divided by .
Solution
The divisors of are ,,,,,,, and , so the problem requires the sum of the number of numbers relatively prime to each of the eight numbers.
Using the formula (or by manually counting the numbers relatively prime to each number), number is relatively prime to , number is relatively prime to , numbers are relatively prime to , numbers are relatively prime to , numbers are relatively prime to , numbers are relatively prime to , numbers are relatively prime to , and numbers are relatively prime to . That means , and the remainder when is divided by is .
Extra Note
If all the divisors of an integer n are {, , , ... }, then n = + + .... + . 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 = .... where are prime numbers
and
{, , .... } is an exhaustive, mutually exclusive list of 's factors
+ + ... +
= + + ...
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.
= + ....
= )) ...
= x
x ...
= ...
=
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 38 |
Followed by: Problem 40 | |
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