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

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~lptoggled
 
~lptoggled
 
==Solution 2==
 
==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>, something which can be easily 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 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 <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>
+
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==
 
==Solution 3==
−
For any real numbers x and y, <math>x^3-y^3=(x-y)(x^2+xy+y^2)=(x-y)((x-y)^2+3xy)=(x-y)^3+3(x-y)xy</math>. When <math>x = y + 1</math>, <math>(x-y)^3+3(x-y)xy</math> becomes <math>1 + 3xy</math>.
+
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>
  
−
Therefore, <cmath> 2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3 = (1 + 3\cdot 1\cdot 2) + (1 + 3\cdot 3\cdot 4) + (1 + 3\cdot 5\cdot 6) + \dots + (1 + 3\cdot 17\cdot 18)</cmath> which is <cmath>\sum_{n=1}^{9}(1 + 3\cdot 2n(2n-1)) = 9 + 3\cdot (1\cdot 2 + 3\cdot 4 + 5\cdot 6 \dots + 17\cdot 18) = 9 + 3 \cdot (1\cdot 2 + 3\cdot 4 + 5\cdot 6 \dots + 17\cdot 18) =\boxed{\textbf{(D) } 3159}</cmath>
+
<math>= \boxed{\textbf{(D) } 3159}</math>
  
 
~sqroot
 
~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) \\
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=(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>
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<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

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