# 2023 AMC 12A Problems/Problem 12

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## Problem

What is the value of $$2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?$$

$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$

## Solution 1

To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas.

$$2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$$

$=(2-1)(2^2+1 \cdot 2+1^2)+(4-3)(4^2+4 \cdot 3+3^2)+(6-5)(6^2+6 \cdot 5+5^2)+...+(18-17)(18^2+18 \cdot 17+17^2)$

$=(2^2+1 \cdot 2+1^2)+(4^2+4 \cdot 3+3^2)+(6^2+6 \cdot 5+5^2)+...+(18^2+18 \cdot 17+17^2)$

$=1^2+2^2+3^2+4^2+5^2+6^2...+17^2+18^2+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$

$=\frac{18(18+1)(36+1)}{6}+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$

we could rewrite the second part as $\sum_{n=1}^{9}(2n-1)(2n)$

$(2n-1)(2n)=4n^2-2n$

$\sum_{n=1}^{9}4n^2=4(\frac{9(9+1)(18+1)}{6})$

$\sum_{n=1}^{9}-2n=-2(\frac{9(9+1)}{2})$

Hence,

$1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18 = 4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$

$2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$

$=\frac{18(18+1)(36+1)}{6}+4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$

$=3(19)(37)+6(10)(19)-9(10)$

$=2109+1140-90$

$=\boxed{\textbf{(D) } 3159}$

~lptoggled

## Solution 2

Think about $2^3+4^3+6^3+...+18^3$. Once we factor out $2^3=8$, we get $1^3+2^3+...+9^3$, which can be found using the sum of cubes formula, $(\frac{n(n+1)}{2})^2$. Now think about $1^3+3^3+...+17^3$. This is just the previous sum subtracted from the total sum of $18$ cubes. So now we have the two things we need to add. The sum of all the even cubes is $8\cdot (\frac{90}{2})^2\rightarrow 8\cdot 2025 = 16200$. The sum of all cubes from $1^3$ to $18^3$ is $(\frac{18\cdot 19}{2})^2=29241$. The sum of the odd cubes is then $29241-16200=13041$. Thus we get $16200-13041=\boxed{\textbf{(D) } 3159}$ ~amcrunner

## Solution 2 (a bit faster)

Using the same sum of cubes formula, we can rewrite as $2(2^3 + 4^3 + ... + 18^3) - (1^3 + 2^3 + ... + 18^3)$

$= 2(2^3)(1^3 + ... + 9^3) - (1^3 + ... + 18^3)$

$= 16(5 \cdot 9)^2 - (9 \cdot 19)^2 = 9^2(20^2 - 19^2) = 81 \cdot 39 = \boxed{\textbf{(D) } 3159}$ ~AoPSuser216

## Solution 3

For any real numbers $x$ and $y$, $x^3-y^3=(x-y)(x^2+xy+y^2)=(x-y)((x-y)^2+3xy)=(x-y)^3+3xy(x-y)$.

When $x = y + 1$, with the above formula, we will get $x^3-y^3=1^3+3xy=1 + 3xy$.

Therefore,

$2^3 - 1^3 + 4^3 - 3^3 + \dots + 18^3 - 17^3$

$= (1 + 3\cdot 1\cdot 2) + (1 + 3\cdot 3\cdot 4) + \dots + (1 + 3\cdot 17\cdot 18)$

$= 9 + 3\cdot (1\cdot 2 + 3\cdot 4 + \dots + 17\cdot 18)$

$= 9 + 3 \cdot (2 + 12 + 30 + 56 + 90 + 132 + 182 + 240 + 306)$

$= 9 + 3 \cdot 1050$

$= \boxed{\textbf{(D) } 3159}$

~sqroot

Alternatively, to avoid the long sum,

$(1\cdot 2 + 3\cdot 4 + \dots + 17\cdot 18) \\ = (2^2)(9(10)(19)/6) - (2)(9(10)/2) \\ =(2)(9)(10)(2(19/6) - 1/2) \\ = 180(35/6) = 35(30) = 1050$

## Solution 4

We rewrite the sum as

$$\sum_{k=1}^{9}(2k)^3-(2k-1)^3$$ $$=\sum_{k=1}^{9} 12k^2 - 6k + 1$$ $$=12\sum_{k=1}^{9}k^2 - 6\sum_{k=1}^{9}k + 9$$ $$=12\cdot\frac{9\cdot 10\cdot 19}{6} -6\cdot \frac{9\cdot 10}{2} + 9$$ $$=3420 - 270 + 9 = \boxed{\textbf{(D) } 3159}$$

-Benedict T (countmath1)

## Solution 4 (a bit faster)

We see $=12\cdot\frac{9\cdot 10\cdot 19}{6} -6\cdot \frac{9\cdot 10}{2} + 9 = 9\cdot (12\cdot\frac{10\cdot 19}{6} -6\cdot \frac{10}{2} + 1)$ which is clearly a multiple of 9. The only answer choice which is a multiple of 9 is $\boxed{\textbf{(D) } 3159}$ ~Ilaggo2432

## Solution 5 (Bash)

$2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3$

$=8-1+64-27+216-125+512-343+1000-729+1728-1331+2744-2197+4096-3375+5832-4913$

$= \boxed{\textbf{(D) } 3159}$

## Solution 6

Reduce all terms mod 9. This yields:

$2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3$ $\equiv -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 +0 - -1 (\mod 9)$ $\equiv 0 (\mod 9)$

The only answer choice which is also ≡0 mod 9 is $= \boxed{\textbf{(D) } 3159}$

## Video Solution

https://youtu.be/33Tz-bfKzmw ~Education, the Study of Everything

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)