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

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==Solution 3==
 
==Solution 3==
  
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We first note that <math>1^3+2^3+...=(1+2+...)^2</math>. So what we are trying to find is what <math>((</math>2018<math>)^(</math>2018<math>))^2=(2018)^(4036)</math> is mod <math>6</math>. We start by noting that <math>2018</math> is congruent to <math>2</math> mod <math>6</math>. So we are trying to find <math>2^4036</math> mod <math>6</math>. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start small powers of <math>2</math> and see that <math>2^1</math> is <math>2</math> mod <math>6</math>, <math>2^2</math> is <math>4</math> mod <math>6</math>, <math>2^3</math> is <math>2</math> mod <math>6</math>, <math>2^4</math> is <math>4</math> mod <math>6</math>, and so on... So we see that since <math>2^(4036)</math> has an even power, it must be congruent to <math>4</math> mod <math>6</math>, thus giving our answer <math>\boxed{\text{(E) }4}</math>. You can prove this pattern using mods. But I thought this was easier.
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We first note that <math>1^3+2^3+...=(1+2+...)^2</math>. So what we are trying to find is what <math>\left(2018^{2018}\right)^3=</math>\left(2018)^{4036}\right<math> is mod </math>6<math>. We start by noting that </math>2018<math> is congruent to </math>2<math> mod </math>6<math>. So we are trying to find </math>2^4036<math> mod </math>6<math>. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start small powers of </math>2<math> and see that </math>2^1<math> is </math>2<math> mod </math>6<math>, </math>2^2<math> is </math>4<math> mod </math>6<math>, </math>2^3<math> is </math>2<math> mod </math>6<math>, </math>2^4<math> is </math>4<math> mod </math>6<math>, and so on... So we see that since </math>2^(4036)<math> has an even power, it must be congruent to </math>4<math> mod </math>6<math>, thus giving our answer </math>\boxed{\text{(E) }4}$. You can prove this pattern using mods. But I thought this was easier.
  
 
-TheMagician
 
-TheMagician

Revision as of 20:43, 16 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

$n^{3}\equiv n \pmod{6}$

Therefore the answer is congruent to $2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}$ Please don't take credit, thanks!

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 $\left(2018^{2018}\right)^3=$\left(2018)^{4036}\right$is mod$6$. We start by noting that$2018$is congruent to$2$mod$6$. So we are trying to find$2^4036$mod$6$. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start 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$2^(4036)$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

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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