Difference between revisions of "2018 AMC 10B Problems/Problem 16"

Revision as of 14:04, 19 February 2018

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

By Euler's Totient Theorem, $n^{3}\equiv n \pmod{6}$ Alternatively, one could simply list out all the residues to the third power $\mod 6$

Therefore the answer is congruent to $2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}$

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 $\left(2018^{2018}\right)^2=\left(2018^{4036}\right)$ mod $6$ . We start by noting that $2018$ is congruent to $2$ mod $6$ . So we are trying to find $\left(2^{4036}\right)$ 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 $2$ 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 $\left(2^{4036}\right)$ 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

Solution 4 (Lazy solution)

Assume $a_1, a_2, ... a_2017$ are multiples of 6 and find $2018^{2018} mod 6$ (which happens to be 4). Then ${a_1}^3 + ... + {a_2018}^3$ is congruent to $64 mod 6$ or just $4$ .

Note: It is pretty easy to find this construction; one example is $a_1, a_2, ... a_{2016} = 6, a_{2018} = 4$ , and then let $a_{2017}$ be whatever is left that you need to add to get $2018^{2018}$ .

-Patrick4President

@@ Line 28: / Line 28: @@
 -TheMagician
+==Solution 4 (Lazy solution)==
+Assume <math>a_1, a_2, ... a_2017</math> are multiples of 6 and find <math>2018^{2018} mod 6</math> (which happens to be 4). Then <math>{a_1}^3 + ... + {a_2018}^3</math> is congruent to <math>64 mod 6</math> or just <math>4</math>.
+Note: It is pretty easy to find this construction; one example is <math>a_1, a_2, ... a_{2016} = 6, a_{2018} = 4</math>, and then let <math>a_{2017}</math> be whatever is left that you need to add to get <math>2018^{2018}</math>.
+-Patrick4President
 ==See Also==

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