Difference between revisions of "2018 AMC 10B Problems/Problem 16"
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-Patrick4President | -Patrick4President | ||
+ | ==Solution 5 (Fermat's Little Theorem)== | ||
+ | |||
+ | First note that each <math>a_{i}^3 \equiv a_{i} \pmod 3</math> by Fermat's Little Theorem. This implies that <math>a_{1}^3+...+a_{2018}^3 \equiv a_{1}+...+a_{2018} \equiv 2^{2018} \equiv 1 \pmod{3}</math>. Also, all <math>a_{i}^2 \equiv a_{i} \pmod{2}</math>, hence <math>a_{i}^3 \equiv (a_{i})(a_{i}^2) \equiv a_{i}^2 \equiv a_{i} \pmod{2}</math> by Fermat's Little Theorem.Thus, <math>a_{1}^3+...a_{2018}^3 \equiv 2^{2018} \equiv 0 \pmod{2}</math>. Now set <math>x=a_{1}^3+...+a_{2018}^3</math>. Then, we have the congruences <math>x \equiv 0 \pmod{2}</math> and <math>x \equiv 1 \pmod{3}</math>. By the Chinese Remainder Theorem, a solution must exist, and indeed solving the congruence we get that <math>x \equiv 4 \pmod{6}</math>. Thus, the answer is <math>\boxed{ (E)4}</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}} | {{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:53, 13 May 2018
Let be a strictly increasing sequence of positive integers such that
What is the remainder when
is divided by
?
Contents
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
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
Solution 4 (Lazy solution)
Assume are multiples of 6 and find
(which happens to be 4). Then
is congruent to
or just
.
-Patrick4President
Solution 5 (Fermat's Little Theorem)
First note that each by Fermat's Little Theorem. This implies that
. Also, all
, hence
by Fermat's Little Theorem.Thus,
. Now set
. Then, we have the congruences
and
. By the Chinese Remainder Theorem, a solution must exist, and indeed solving the congruence we get that
. Thus, the answer is
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.