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

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-Patrick4President
 
-Patrick4President
  
==Solution 5 (Fermat's Little Theorem)==
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==Solution 6 (Nichomauss'Theorem)==
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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}^2.
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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}$
  
First note that each <math>a_{i}^3 \equiv a_{i} \pmod 3</math> by Fermat's Little Theorem. This implies that <math>a_{1}^3+...+a_{2018}^3 \equiv a_{1}+...+a_{2018} \equiv 2^{2018} \equiv 1 \pmod{3}</math>. Also, all <math>a_{i}^2 \equiv a_{i} \pmod{2}</math>, hence <math>a_{i}^3 \equiv (a_{i})(a_{i}^2) \equiv a_{i}^2 \equiv a_{i} \pmod{2}</math> by Fermat's Little Theorem.Thus, <math>a_{1}^3+...a_{2018}^3 \equiv 2^{2018} \equiv 0 \pmod{2}</math>. Now set <math>x=a_{1}^3+...+a_{2018}^3</math>. Then, we have the congruences <math>x \equiv 0 \pmod{2}</math> and <math>x \equiv 1 \pmod{3}</math>. By the Chinese Remainder Theorem, a solution must exist, and indeed solving the congruence we get that <math>x \equiv 4 \pmod{6}</math>. Thus, the answer is <math>\boxed{ (E)4}</math>
 
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}}
 
{{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:31, 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}^2.

Now, we find$ (Error compiling LaTeX. ! Double superscript.)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$\boxed{\textbf{(E)} 4}$

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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