Difference between revisions of "2018 AMC 10B Problems/Problem 16"
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Revision as of 20:31, 24 January 2021
Let be a strictly increasing sequence of positive integers such that What is the remainder when is divided by ?
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.
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 solution, 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.
Solution 4 (Lazy solution)
First, we can assume that the problem will have a consistent answer for all possible values of . For the purpose of this solution, assume are multiples of 6 and find (which happens to be ). Then is congruent to or just .
Solution 5 (Nichomauss' Theorem)
Seeing the cubes of numbers, we think of Nichomauss's theorem, which states that . We can do this and deduce that squared.
Now, we find , which is 2. This means that we need to find , which we can find using a pattern to be . Therefore, the answer is , which is congruent to
Minor edits by fasterthanlight
Algebraic Insight into Given Property
Mods is a good way to prove : residues are simply . Only and are necessary to check. Another way is to observe that factors into . Any consecutive numbers must be a multiple of , so is both divisible by and . This provides an algebraic method for proving for all .
Video Solution 1
With Modular Arithmetic Intro https://www.youtube.com/watch?v=wbv3TArroSs
Video Solution 2
Video Solution 3
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