2018 AMC 10B Problem 16

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

n^3 \equiv n \pmod 6,

because

n^3 - n = (n-1)n(n+1)

is a product of three consecutive integers

So the sum of cubes is congruent to

2018^{2018} \pmod 6;

reduce that modulo

6

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.

Problem 16 in other years