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Difference between revisions of "2018 AMC 10B Problems/Problem 16"
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(not very good one)
 
(not very good one)
  
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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\prod_{i\neq j\neq k}^{2018} a_ia_ja_k</math>
+
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\prod_{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})
+
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 11:14, 3 August 2018

Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution 1

One could simply list out all the residues to the third power $\mod 6$. (Edit: Euler's totient theorem is not a valid approach to showing that they are all congruent $\mod 6$. This is due to the fact that $a_k$ need not be relatively prime to $6$.)

Therefore the answer is congruent to $2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}$

Solution 2

(not very good one)

Note that $\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k$

Note that $a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2(2018-a_1)+3a_2^2(2018-a_2)+\cdots+3a_{2018}^2(2018-a_{2018}) \equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6$ Therefore, $-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}$.

Thus, $a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3$. 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 $\boxed{\text{(E) }4}$

Solution 3

We first note that $1^3+2^3+...=(1+2+...)^2$. So what we are trying to find is what $\left(2018^{2018}\right)^2=\left(2018^{4036}\right)$ mod $6$. We start by noting that $2018$ is congruent to $2$ mod $6$. So we are trying to find $\left(2^{4036}\right)$ mod $6$. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with small powers of $2$ and see that $2^1$ is $2$ mod $6$, $2^2$ is $4$ mod $6$, $2^3$ is $2$ mod $6$, $2^4$ is $4$ mod $6$, and so on... So we see that since $\left(2^{4036}\right)$ has an even power, it must be congruent to $4$ mod $6$, thus giving our answer $\boxed{\text{(E) }4}$. You can prove this pattern using mods. But I thought this was easier.

-TheMagician

Solution 4 (Lazy solution)

Assume $a_1, a_2, ... a_{2017}$ are multiples of 6 and find $2018^{2018} mod 6$ (which happens to be 4). Then ${a_1}^3 + ... + {a_{2018}}^3$ is congruent to $64 mod 6$ or just $4$.

-Patrick4President

Solution 5 (Fermat's Little Theorem)

First note that each $a_{i}^3 \equiv a_{i} \pmod 3$ by Fermat's Little Theorem. This implies that $a_{1}^3+...+a_{2018}^3 \equiv a_{1}+...+a_{2018} \equiv 2^{2018} \equiv 1 \pmod{3}$. Also, all $a_{i}^2 \equiv a_{i} \pmod{2}$, hence $a_{i}^3 \equiv (a_{i})(a_{i}^2) \equiv a_{i}^2 \equiv a_{i} \pmod{2}$ by Fermat's Little Theorem.Thus, $a_{1}^3+...a_{2018}^3 \equiv 2^{2018} \equiv 0 \pmod{2}$. Now set $x=a_{1}^3+...+a_{2018}^3$. Then, we have the congruences $x \equiv 0 \pmod{2}$ and $x \equiv 1 \pmod{3}$. By the Chinese Remainder Theorem, a solution must exist, and indeed solving the congruence we get that $x \equiv 4 \pmod{6}$. Thus, the answer is $\boxed{ (E)4}$

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

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