Difference between revisions of "2023 AMC 12A Problems/Problem 12"
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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>, 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> | ||
~amcrunner | ~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*9)^2 - (9*19)^2 = 9^2(20^2 - 19^2) = 81*39 = \boxed{\textbf{(D) } 3159}</math> | ||
+ | ~AoPSuser216 | ||
==Solution 3== | ==Solution 3== |
Revision as of 16:55, 10 November 2023
Contents
[hide]Problem
What is the value of
Solution 1
To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas.
we could rewrite the second part as
Hence,
Adding everything up:
~lptoggled
Solution 2
Think about . Once we factor out , we get , something which can be easily found using the sum of cubes formula, . Now think about . 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 . The sum of all cubes from to is . The sum of the odd cubes is then . Thus we get ~amcrunner
Solution 2 (a bit faster)
Using the same sum of cubes formula, we can rewrite as
~AoPSuser216
Solution 3
For any real numbers and , .
When , with the above formula, we will get .
Therefore,
~sqroot
Solution 4
We rewrite the sum as
-Benedict T (countmath1)
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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