Difference between revisions of "2018 AMC 10B Problems/Problem 16"
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==Solution 1== | ==Solution 1== | ||
− | <math>n^{3}\equiv n \pmod{6}</math> | + | By [[Euler's Totient Theorem]], <math>n^{3}\equiv n \pmod{6}</math> |
Therefore the answer is congruent to <math>2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}</math> | Therefore the answer is congruent to <math>2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}</math> |
Revision as of 19:04, 17 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
Solution 2
(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 3
We first note that . So what we are trying to find is what
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 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
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.