2018 AMC 10B Problems/Problem 16
Contents
Problem
Let be a strictly increasing sequence of positive integers such that What is the remainder when is divided by ?
Solution 1
One could simply list out all the residues to the third power . (Edit: Euler's totient theorem is not a valid approach to showing that they are all congruent . This is due to the fact that need not be relatively prime to .)
Therefore the answer is congruent to
Note from Williamgolly: we can wlog assume and have to make life easier
Solution 2
Note that
Note that Therefore, .
Thus, . However, since cubing preserves parity, and the sum of the individual terms is even, the some of the cubes is also even, and our answer is
Solution 3 (Partial Proof)
First, we can assume that the problem will have a consistent answer for all possible values of . For the purpose of this problem, we will assume that
We first note that . So what we are trying to find is what mod . We start by noting that is congruent to . So we are trying to find . Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with small powers of and see that is mod , is mod , is mod , is mod , and so on... So we see that since has an even power, it must be congruent to , thus giving our answer . You can prove this pattern using mods. But I thought this was easier.
-TheMagician
Solution 4 (Lazy solution)
Assume are multiples of 6 and find (which happens to be ). Then is congruent to or just .
-Patrick4President
Solution 5 (Fermat's Little Theorem)
First note that each by Fermat's Little Theorem. This implies that . Also, all , hence by Fermat's Little Theorem.Thus, . Now set . Then, we have the congruences and . By the Chinese Remainder Theorem, a solution must exist, and indeed solving the congruence we get that . Thus, the answer is
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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All AMC 10 Problems and Solutions |
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