All About 2018 -- B

Number Theory Level 3

Let $a_1,a_2, a_3 \cdots a_{2018}$ be a strictly increasing sequence of positive integers such that

$\large a_1+a_2+a_3+\cdots+a_{2018}=2018^{2018}$

What is the remainder when $a_1^3+a_2^3+a_3^3\cdots+a_{2018}^3$ is divided by 6?

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1 solution

Pierre Stöber
Jun 8, 2018

Notice that for every integer n, the following is true:

n mod 2 =n^3 mod 2

n mod 3 =n^3 mod 3

Then, this is also true:

n mod 6 =n^3 mod 6

This means we can get rid of the exponents. Consequentially, the answer is equal to the first sum modulo 6. Then, we get:

2018^2018 mod 6 = 2^2 mod 6 = 4 (we used the fact that 2018 modulo 6= 2)