2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 16 Medium

Let a_1, a_2, \cdots, 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? (2018 AMC 10B Problem, Question#16)

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    3

  • E.

    4

Answer:E

One could simply list out all the residues to the third power \rm mod~6. (Edit: Euler's totient theorem is not a valid approach to showing that they are all congruent \rm 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}\equiv2^{2018} (\rm mod \ 6)=(E) \ 4.

Note that

(a_1+a_2+\cdots+a_{2018})^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2(a_1+a_2+\cdots+a_{2018}-a_1)+3a_2^2(a_1+a_2+\cdots+a_{2018}-a_2)+\cdots+3a_{2018}^2(a_1+a_2+\cdots+a_{2018}-a_{2018})+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(a_1+a_2+\cdots+a_{2018}-a_1)+3a_2^2(a_1+a_2+\cdots+a_{2018}-a_2)+\cdots+3a_{2018}^2(a_1+a_2+\cdots+a_{2018}-a_{2018})+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^{2018}-a_1)+3a_2^2(2018^{2018}-a_2)+\cdots+3a_{2018}^2(2018^{2018}-a_{2018})\equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)(\rm mod ~6),

Therefore,-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv(2018^{2018})^3\equiv(2^{2018})^3\equiv4^3\equiv4(\rm mod ~6).

Thus, a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1(\rm mod ~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 \rm (\rm E)~4.

We first note that 1^3+2^3+\cdots=(1+2+\cdots)^2. So what we are trying to find is what (2018^{2018})^2=(2018^{4036}) \ \rm mod \ 6. We start by noting that 2018 is congruent to 2 \ \rm mod \ 6. So we are trying to find (2^{4036}) \ \rm 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 \ \rm mod \ 6,2^2  is 4 \ \rm mod \ 6 , 2^3 is 2 \ \rm mod \ 6, 2^4 is 4 \ \rm mod \ 6 ,  and so on \cdots So we see that since (2^{4036}) has an even power, it must be congruent to 4 \ \rm mod \ 6 , thus giving our answer \rm (E) \ 4.

You can prove this pattern using mods. But I thought this was easier.

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

First note that each a_i^3\equiv a_i(\rm mod \ 3) by Fermat's Little Theorem. This implies that a_1^3+\cdots+a_{2018}^3\equiv a_1+\cdots+a_{2018}\equiv 2^{2018}\equiv 1(\rm mod \ 3). Also, all  a_i^2\equiv a_i(\rm mod \ 2), hence a_i^3\equiv(a_i)(a_i^2)\equiv a_i^2\equiv a_i(\rm mod \ 2) by Fermat's Little, Theorem.Thus, a_1^3+\cdots+a_{2018}^3\equiv 2^{2018}\equiv 0(\rm mod \ 2). Now set x=a_1^3+\cdots +a_{2018}^3 . Then, we have the congruences x\equiv 0(\rm mod \ 2) and x\equiv 1(\rm mod \ 3) . By the Chinese Remainder Theorem, a solution must exist, and indeed solving the congruence we get that x\equiv 4(\rm mod \ 6). Thus, the answer is \rm (E) \ 4.

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