Difference between revisions of "2018 AMC 10B Problems/Problem 16"
TheMagician (talk | contribs) (→Solution 2) |
TheMagician (talk | contribs) (→Solution) |
||
Line 10: | Line 10: | ||
Therefore the answer is congruent to <math>2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}</math> Please don't take credit, thanks! | Therefore the answer is congruent to <math>2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}</math> Please don't take credit, thanks! | ||
− | == | + | ==Solution2== |
(not very good one) | (not very good one) | ||
Revision as of 21:40, 16 February 2018
Let be a strictly increasing sequence of positive integers such that What is the remainder when is divided by ?
Contents
Solution 1
Therefore the answer is congruent to Please don't take credit, thanks!
Solution2
(not very good one)
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 2
We first note that . So what we are trying to find is what 20182018 is mod . We start by noting that is congruent to mod . So we are trying to find mod . Instead of trying to do this with some number theory skills, we could just look for a pattern. We start small powers of and see that is mod , is mod , is mod , is mod , and so on... So we see that since $2^$ (Error compiling LaTeX. ! Missing { inserted.) has an even power, it must be congruent to mod , thus giving our answer . You can prove this pattern using mods. But I thought this was easier.
-TheMagician
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.