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

Revision as of 11:11, 4 January 2024

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

Note that, because of Fermat's Little Theorem, for any integer $a$ relatively prime to 6, we have $a^2 \equiv 1 \pmod{6} \implies a^3 \equiv a \pmod{6}$ . This is also trivially true if $a$ is not relatively prime to 6 because $a \equiv 0 \pmod{6}$ . Therefore,

$a_1+a_2+\cdots+a_{2018} \equiv a_1^3+a_2^3+\cdots+a_{2018}^3 \pmod{6}.$

Therefore the answer is congruent to $2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ \text{(E)}4}$ because $2^n \pmod{6}$ alternates with $2$ and $4$ when $n$ increases.

~Dolphindesigner

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 sum 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 $\boxed{\textbf{(E)} 4}$ .

-Patrick4President

~minor edit made by CatachuKetchup~

Solution 5 (Even Lazier Solution)

Due to the large amounts of variables in the problem, and the fact that the test is only 75 minutes, you can assume that the answer is probably just $2018^{2018} \pmod{6}$ , which is $\boxed{\textbf{(E)} 4}$ .

~ Zeeshan12 [Be warned that this technique is not recommended for all problems and you should use it as a last resort]

Algebraic Insight into Given Property

Mods is a good way to prove $a^3 \equiv a \pmod6$ : residues are simply $3, \pm 2, \pm 1, 0$ . Only $2$ and $3$ are necessary to check. Another way is to observe that $a^3-a$ factors into $(a-1)a(a+1)$ . Any $k$ consecutive numbers must be a multiple of $k$ , so $a^3-a$ is both divisible by $2$ and $3$ . This provides an algebraic method for proving $a^3 \equiv a \pmod6$ for all $a$ .

Video Solution 1

With Modular Arithmetic Intro https://www.youtube.com/watch?v=wbv3TArroSs

~IceMatrix

Video Solution 2

https://www.youtube.com/watch?v=SRjZ6B5DR74

~bunny1

Video Solution 3 by OmegaLearn

https://youtu.be/4_x1sgcQCp4?t=112

~ pi_is_3.14

@@ Line 18: / Line 18: @@
 ==Solution 2==
-Note that <math>\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</math>
+Note that <math>\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</math>
 Note that <math>

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