Difference between revisions of "2018 AMC 10B Problems/Problem 16"
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==Solution 5 (Even Lazier Solution)== | ==Solution 5 (Even Lazier Solution)== | ||
− | Due to the large amounts of variables in the problem, and the fact that the test is only 75 minutes, you can assume that the answer is probably just <math>2018^{2018} | + | Due to the large amounts of variables in the problem, and the fact that the test is only 75 minutes, you can assume that the answer is probably just <math>2018^{2018} (mod 6) </math>, which is <math>\boxed{\textbf{(E)} 4}</math>. |
− | ~ Zeeshan12 | + | ~ Zeeshan12 [Be warned that this technique is not recommended for all problems and you should use it as a last resort (or when you intuitively know to use it)] |
− | [Be warned that this technique is not recommended for all problems and you should use it as a last resort] | ||
==Algebraic Insight into Given Property== | ==Algebraic Insight into Given Property== |
Revision as of 16:32, 30 May 2022
Contents
[hide]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 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.
-TheMagician
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 .
-Patrick4President
~minor edit made by CatachuKetchup~
Solution 5 (Even Lazier Solution)
Due to the large amounts of variables in the problem, and the fact that the test is only 75 minutes, you can assume that the answer is probably just , which is .
~ Zeeshan12 [Be warned that this technique is not recommended for all problems and you should use it as a last resort (or when you intuitively know to use it)]
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
~IceMatrix
Video Solution 2
https://www.youtube.com/watch?v=SRjZ6B5DR74
~bunny1
Video Solution 3
https://youtu.be/4_x1sgcQCp4?t=112
~ pi_is_3.14
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.