Mock AIME 3 Pre 2005 Problems/Problem 10
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Problem
is a sequence of positive integers such that
for all integers . Compute the remainder obtained when is divided by if .
Solution
We can easily express as the following sum:
This can be broken down into three simpler sums:
Using finite calculus, one can easily derive the following classical sums:
Using these, we can now compute:
And hence we get the closed form for :
Now we can compute as follows:
We know that (where is the Euler's totient function), hence , and thus . Thus there is an integer such that . And then , hence .
Therefore we have .
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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