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

(Solution 2)
(Solution 2)
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==Solution 2==
 
==Solution 2==
  
We first note that <math>1^3+2^3+...=(1+2+...)^2</math>. So what we are trying to find is what <math>((2018^2018))^2</math><math>=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>((2018^2018)</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.
  
 
-TheMagician
 
-TheMagician

Revision as of 21:36, 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

(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 2

We first note that $1^3+2^3+...=(1+2+...)^2$. So what we are trying to find is what $((2018^2018)$ 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 (ProblemsAnswer KeyResources)
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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