2018 AMC 10B Problems/Problem 16

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

$n^{3}\equiv n \pmod{6}$

Therefore the answer is congruent to $2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}$ Please don't take credit, thanks!

Solution 2

(not very good one)

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

Note that $a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\prod_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2(2018-a_1)+3a_2^2(2018-a_2)+\cdots+3a_{2018}^2(2018-a_{2018}) \equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6$ 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 $\left(2018^{2018}\right)^3=$ \left(2018)^{4036}\right $is mod$ 6 $. We start by noting that$ 2018 $is congruent to$ 2 $mod$ 6 $. So we are trying to find$ 2^4036 $mod$ 6 $. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start small powers of$ 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 $, and so on... So we see that since$ 2^(4036) $has an even power, it must be congruent to$ 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

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2018 AMC 10B Problems/Problem 16

Contents

Solution 1

Solution 2

Solution 3

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