Difference between revisions of "2018 AMC 10B Problems/Problem 16"
Ericshi1685 (talk | contribs) (→Solution 6 (Nichomauss' Theorem)) |
Coltsfan10 (talk | contribs) (→Solution 1) |
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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> | ||
− | Note from Williamgolly: we can wlog assume <math>a_1,a_2... | + | Note from Williamgolly: we can wlog assume <math>a_1,a_2... a_(2017)=0(mod6)</math> and have <math>a_2018=2(mod6)</math> to make life easier |
==Solution 2== | ==Solution 2== |
Revision as of 13:31, 3 May 2020
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 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
Solution 6 (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
-ericshi1685
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.