Difference between revisions of "2018 AMC 10B Problems/Problem 16"
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− | Note that <math>\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\ | + | Note that <math>\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k</math> |
Note that <math> | Note that <math> | ||
− | a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\ | + | a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2(2018-a_1)+3a_2^2(2018-a_2)+\cdots+3a_{2018}^2(2018-a_{2018}) |
\equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6 | \equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6 | ||
</math> | </math> |
Revision as of 12:14, 3 August 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.