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

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Seeing the cubes of numbers, we think of Nichomauss's theorem, which states that <math>(a_1^3 + a_2^3 + ... + a_n^3) = (a_1 + a_2 + ... + a_n)^2</math>. We can do this and deduce that <math>(a_1^3 + a_2^3 + ... + a_{2018}^3) = 2018^{2018}</math> squared.  
 
Seeing the cubes of numbers, we think of Nichomauss's theorem, which states that <math>(a_1^3 + a_2^3 + ... + a_n^3) = (a_1 + a_2 + ... + a_n)^2</math>. We can do this and deduce that <math>(a_1^3 + a_2^3 + ... + a_{2018}^3) = 2018^{2018}</math> squared.  
  
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Now, we find <math>2018 \mod 6</math>, which is 2. This means that we need to find <math>2^{2018} \mod {6}</math>, which we can find using a pattern to be <math>4</math>. Therefore, the answer is <math>4^3 \mod 6</math>, which is congruent to <math>\boxed{\textbf{(E)} 4}</math>
+
Now, we find <math>2018\mod 6</math>, which is 2. This means that we need to find <math>2^{2018} \mod {6}</math>, which we can find using a pattern to be <math>4</math>. Therefore, the answer is <math>4^3\mod 6</math>, which is congruent to <math>\boxed{\textbf{(E)} 4}</math>
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-ericshi1685
  
 
==See Also==
 
==See Also==

Revision as of 15:19, 27 March 2020

Problem

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

One could simply list out all the residues to the third power $\mod 6$. (Edit: Euler's totient theorem is not a valid approach to showing that they are all congruent $\mod 6$. This is due to the fact that $a_k$ need not be relatively prime to $6$.)

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

Note from Williamgolly: we can wlog assume $a_1,a_2... a_2017=0(mod6)$ and have $a_2018=2(mod6)$ to make life easier

Solution 2

Note that $\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k$

Note that $a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2({2018}^{2018}-a_1)+3a_2^2({2018}^{2018}-a_2)+\cdots+3a_{2018}^2({2018}^{2018}-a_{2018}) \equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6$ 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 (Partial Proof)

First, we can assume that the problem will have a consistent answer for all possible values of $a_1$. For the purpose of this solution, we will assume that $a_1 = 1$.

We first note that $1^3+2^3+...+n^3 = (1+2+...+n)^2$. 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 $2 \pmod{6}$. So we are trying to find $\left(2^{4036}\right) \pmod{6}$. 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 $4 \pmod{6}$, 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)

First, we can assume that the problem will have a consistent answer for all possible values of $a_1$. For the purpose of this solution, assume $a_1, a_2, ... a_{2017}$ are multiples of 6 and find $2018^{2018} \pmod{6}$ (which happens to be $4$). Then ${a_1}^3 + ... + {a_{2018}}^3$ is congruent to $64 \pmod{6}$ or just $4$.

-Patrick4President

Solution 6 (Nichomauss' Theorem)

Seeing the cubes of numbers, we think of Nichomauss's theorem, which states that $(a_1^3 + a_2^3 + ... + a_n^3) = (a_1 + a_2 + ... + a_n)^2$. We can do this and deduce that $(a_1^3 + a_2^3 + ... + a_{2018}^3) = 2018^{2018}$ squared.

Now, we find $2018\mod 6$, which is 2. This means that we need to find $2^{2018} \mod {6}$, which we can find using a pattern to be $4$. Therefore, the answer is $4^3\mod 6$, which is congruent to $\boxed{\textbf{(E)} 4}$

-ericshi1685

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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