Difference between revisions of "2023 AMC 12A Problems/Problem 12"

Revision as of 19:07, 10 November 2023

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,

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

Adding everything up:

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

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

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

$=2109+1140-90$

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

~lptoggled

Solution 2

Think about $2^3+4^3+6^3+...+18^3$ . Once we factor out $2^3=8$ , we get $1^3+2^3+...+9^3$ , something which can be easily 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. We just need to not screw up the computations: the sum of all the even cubes is just $8\cdot (\frac{90}{2})^2\rightarrow 8\cdot 2025 = 16200$ . 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 $2(2^3 + 4^3 + ... + 18^3) - (1^3 + 2^3 + ... + 18^3)$

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

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

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)

Video Solution

https://youtu.be/_eoPL5H8b0k

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

@@ Line 88: / Line 88: @@
 -Benedict T (countmath1)
+==Video Solution==
+https://youtu.be/_eoPL5H8b0k
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 ==See also==
 {{AMC12 box|year=2023|ab=A|num-b=11|num-a=13}}
 {{MAA Notice}}

2023 AMC 12A (Problems • Answer Key • Resources)
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