Question

Let $a_1, a_2, \ldots, a_{2018}$ 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 $6?$

$0$

$1$

$2$

$3$

$4$

Answer: E

Concepts:modular arithmetic divisibility

Difficulty rating: 1710

Solution:

For any integer $n,$ $n^3 - n = (n-1)n(n+1)$ is a product of three consecutive integers, so it's divisible by $6.$ That means $n^3 \equiv n \pmod 6.$ Summing, $\sum a_i^3 \equiv \sum a_i$ $= 2018^{2018} \pmod 6.$ Now $2018 \equiv 2 \pmod 6,$ and powers of $2$ mod $6$ alternate $2, 4, 2, 4, \ldots.$ The exponent $2018$ is even, so $2^{2018} \equiv 4 \pmod 6.$ The remainder is $4.$ Therefore, the answer is E.

2018 AMC 10B Problem 16

Answer: E

Difficulty rating: 1710

Solution:

Problem 16 in Other Years