Difference between revisions of "2023 AMC 12A Problems/Problem 12"
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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 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>. | ||
− | 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 ( | + | 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 (2 + 12 + 30 + 56 + 90 + 132 + 182 + 240 + 306) =\boxed{\textbf{(D) } 3159}</cmath> |
~sqroot | ~sqroot |
Revision as of 03:11, 10 November 2023
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
Solution 3
For any real numbers x and y, . When
,
becomes
.
Therefore, which is
~sqroot
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.