Difference between revisions of "2021 AMC 12A Problems/Problem 9"
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~mathboy100 | ~mathboy100 | ||
+ | |||
+ | == Solution 6(Generalization) == | ||
+ | Let’s generalize the numbers 2 and 3 to variables <math>y</math> and <math>x</math>. Then we get: | ||
+ | <cmath>(y+x)(y^2+x^2)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64})</cmath> | ||
+ | |||
+ | We see that the first term is <math>y+x</math>, and the next is <math>y^2+x^2</math>. We realize that if we multiply the first term by <math>y-x</math>, the result by difference of squares will be <math>y^2-x^2</math>: we can proceed to use difference of squares on that. In other words, the equation has the domino effect, and all we need to get started is to multiply the whole equation by <math>y-x</math> (keeping in mind to divide <math>y-x</math> at the end.) | ||
+ | |||
+ | Now we get: | ||
+ | |||
+ | <cmath>(y-x)(y+x)(y^2+x^2)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64})</cmath> | ||
+ | <cmath>=(y^2-x^2)(y^2+x^2)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64})</cmath> | ||
+ | <cmath>=(y^4-x^4)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64})</cmath> | ||
+ | <cmath>\cdots</cmath> | ||
+ | <cmath>=(y^{64}-x^{64})(y^{64}+x^{64}) = y^{128}-x^{128}</cmath> | ||
+ | |||
+ | Now we can plug in <math>y = 2</math> and <math>x = 3</math>: | ||
+ | |||
+ | <cmath>(2-3)(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64}) = 2^{128}-3^{128}</cmath>. | ||
+ | |||
+ | However, we must not forget to divide by <math>2-3 = -1</math> at the end! Dividing, we get the answer of | ||
+ | <cmath>3^{128}-2^{128} = \boxed{C}.</cmath> | ||
+ | |||
+ | -KingRavi | ||
+ | |||
==Video Solution== | ==Video Solution== |
Revision as of 03:10, 14 March 2021
- The following problem is from both the 2021 AMC 10A #10 and 2021 AMC 12A #9, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (Engineer's Induction)
- 6 Solution 5 (Difference of Squares)
- 7 Solution 6(Generalization)
- 8 Video Solution
- 9 Video Solution by Aaron He
- 10 Video Solution(Conjugation, Difference of Squares)
- 11 Video Solution by Hawk Math
- 12 Video Solution by OmegaLearn(Factorizations/Telescoping& Meta-solving)
- 13 Video Solution 6
- 14 Video Solution by TheBeautyofMath
- 15 See also
Problem
Which of the following is equivalent to
Solution 1
All you need to do is multiply the entire equation by . Then all the terms will easily simplify by difference of squares and you will get
or
as your final answer. Notice you don't need to worry about
because that's equal to
.
-Lemonie
Solution 2
If you weren't able to come up with the insight, then you could just notice that the answer is divisible by
, and
. We can then use Fermat's Little Theorem for
on the answer choices to determine which of the answer choices are divisible by both
and
. This is
.
-mewto
Solution 3
After expanding the first few terms, the result after each term appears to be where n is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by
would give
, and all the previous terms multiplied by
would give
. Their sum is equal to
, so the proof is complete. Since
is equal to
, the answer is
.
-SmileKat32
Solution 4 (Engineer's Induction)
We can compute some of the first few partial products, and notice that . As we don't have to prove this, we get the product is
, and smugly click
. ~rocketsri
Solution 5 (Difference of Squares)
We notice that the first term is equal to . If we multiply this by the second term, then we will get
, and we can simplify by using difference of squares to obtain
. If we multiply this by the third term and simplify using difference of squares again, we get
. We can continue down the line until we multiply by the last term,
, and get
.
~mathboy100
Solution 6(Generalization)
Let’s generalize the numbers 2 and 3 to variables and
. Then we get:
We see that the first term is , and the next is
. We realize that if we multiply the first term by
, the result by difference of squares will be
: we can proceed to use difference of squares on that. In other words, the equation has the domino effect, and all we need to get started is to multiply the whole equation by
(keeping in mind to divide
at the end.)
Now we get:
Now we can plug in and
:
.
However, we must not forget to divide by at the end! Dividing, we get the answer of
-KingRavi
Video Solution
-Education the Study of Everything
Video Solution by Aaron He
https://www.youtube.com/watch?v=xTGDKBthWsw&t=9m30s
Video Solution(Conjugation, Difference of Squares)
https://www.youtube.com/watch?v=gXaIyeMF7Qo&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=9
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution by OmegaLearn(Factorizations/Telescoping& Meta-solving)
~ pi_is_3.14
Video Solution 6
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/s6E4E06XhPU?t=771 (for AMC 10A)
https://youtu.be/cckGBU2x1zg?t=548 (for AMC 12A)
~IceMatrix
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 10 Problems and Solutions |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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.