Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
m (→‎Solution 4 (a bit faster))
 
(37 intermediate revisions by 12 users not shown)
Line 19: Line 19:
 
<math>=\frac{18(18+1)(36+1)}{6}+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18</math>
 
<math>=\frac{18(18+1)(36+1)}{6}+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18</math>
  
−
we could rewrite the second part as <math>(2n-1)(2n), (1 \leq n \leq 9)</math>
+
we could rewrite the second part as <math>\sum_{n=1}^{9}(2n-1)(2n)</math>
  
 
<math>(2n-1)(2n)=4n^2-2n</math>
 
<math>(2n-1)(2n)=4n^2-2n</math>
Line 44: Line 44:
  
 
~lptoggled
 
~lptoggled
 +
==Solution 2==
 +
Think about <math>2^3+4^3+6^3+...+18^3</math>. Once we factor out <math>2^3=8</math>, we get <math>1^3+2^3+...+9^3</math>, which can be found using the sum of cubes formula, <math>(\frac{n(n+1)}{2})^2</math>. Now think about <math>1^3+3^3+...+17^3</math>. This is just the previous sum subtracted from the total sum of <math>18</math> cubes. So now we have the two things we need to add. The sum of all the even cubes is <math>8\cdot (\frac{90}{2})^2\rightarrow 8\cdot 2025 = 16200</math>. The sum of all cubes from <math>1^3</math> to <math>18^3</math> is <math>(\frac{18\cdot 19}{2})^2=29241</math>. The sum of the odd cubes is then <math>29241-16200=13041</math>. Thus we get <math>16200-13041=\boxed{\textbf{(D) } 3159}</math>
 +
~amcrunner
 +
 +
==Solution 2 (a bit faster)==
 +
Using the same sum of cubes formula, we can rewrite as <math>2(2^3 + 4^3 + ... + 18^3) - (1^3 + 2^3 + ... + 18^3)</math>
 +
 +
<math>= 2(2^3)(1^3 + ... + 9^3) - (1^3 + ... + 18^3)</math>
 +
 +
<math>= 16(5 \cdot 9)^2 - (9 \cdot 19)^2 = 9^2(20^2 - 19^2) = 81 \cdot 39 = \boxed{\textbf{(D) } 3159}</math>
 +
~AoPSuser216
 +
 +
==Solution 3==
 +
For any real numbers <math>x</math> and <math>y</math>, <math>x^3-y^3=(x-y)(x^2+xy+y^2)=(x-y)((x-y)^2+3xy)=(x-y)^3+3xy(x-y)</math>.
 +
 +
When <math>x = y + 1</math>, with the above formula, we will get <math>x^3-y^3=1^3+3xy=1 + 3xy</math>.
 +
 +
Therefore,
 +
 +
<math>2^3 - 1^3 + 4^3 - 3^3 + \dots + 18^3 - 17^3</math>
 +
 +
<math>= (1 + 3\cdot 1\cdot 2) + (1 + 3\cdot 3\cdot 4) + \dots + (1 + 3\cdot 17\cdot 18)</math>
 +
 +
<math>= 9 + 3\cdot (1\cdot 2 + 3\cdot 4 + \dots + 17\cdot 18)</math>
 +
 +
<math>= 9 + 3 \cdot (2 + 12 + 30 + 56 + 90 + 132 + 182 + 240 + 306)</math>
 +
 +
<math>= 9 + 3 \cdot 1050</math>
 +
 +
<math>= \boxed{\textbf{(D) } 3159}</math>
 +
 +
~sqroot
 +
 +
Alternatively, to avoid the long sum,
 +
 +
<math> (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</math>
 +
 +
==Solution 4==
 +
We rewrite the sum as
 +
 +
<cmath>\sum_{k=1}^{9}(2k)^3-(2k-1)^3</cmath>
 +
<cmath>=\sum_{k=1}^{9} 12k^2 - 6k + 1</cmath>
 +
<cmath>=12\sum_{k=1}^{9}k^2 - 6\sum_{k=1}^{9}k + 9</cmath>
 +
<cmath>=12\cdot\frac{9\cdot 10\cdot 19}{6} -6\cdot \frac{9\cdot 10}{2} + 9</cmath>
 +
<cmath>=3420 - 270 + 9 = \boxed{\textbf{(D) } 3159}</cmath>
 +
 +
-Benedict T (countmath1)
 +
 +
==Solution 4 (answer choices)==
 +
We see <math>=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)</math> which is clearly a multiple of 9. The only answer choice which is a multiple of 9 is <math>\boxed{\textbf{(D) } 3159}</math> ~Ilaggo2432
 +
 +
==Solution 5 (Bash)==
 +
 +
<math> 2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3</math>
 +
 +
<math>=8-1+64-27+216-125+512-343+1000-729+1728-1331+2744-2197+4096-3375+5832-4913</math>
 +
 +
<math>= \boxed{\textbf{(D) } 3159}</math>
 +
 +
==Solution 6==
 +
 +
Reduce all terms mod 9. This yields:
 +
 +
<math> 2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3</math>
 +
<math> \equiv -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 +0 - -1 (\mod 9)</math>
 +
<math> \equiv 0 (\mod 9)</math>
 +
 +
The only answer choice which is also ≡0 mod 9 is <math>= \boxed{\textbf{(D) } 3159}</math>
 +
 +
==Video Solution ==
 +
 +
by Power Solve
 +
 +
https://youtu.be/YXIH3UbLqK8?si=RZhSDIKjRNLrgVS5&t=1552
 +
 +
==Video Solution==
 +
https://youtu.be/33Tz-bfKzmw
 +
<i> ~Education, the Study of Everything </i>
 +
 +
==Video Solution==
 +
 +
https://youtu.be/_eoPL5H8b0k
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2023|ab=A|num-b=11|num-a=13}}
 
{{AMC12 box|year=2023|ab=A|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 06:26, 24 August 2024

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 $1^3+2^3+...+9^3$, 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 (answer choices)

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

by Power Solve

https://youtu.be/YXIH3UbLqK8?si=RZhSDIKjRNLrgVS5&t=1552

Video Solution

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

Video Solution

https://youtu.be/_eoPL5H8b0k

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

See also

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_12&oldid=225663"