Difference between revisions of "2021 AMC 12A Problems/Problem 9"
Pi is 3.14 (talk | contribs) (→Video Solution by pi_is_3.14(Factorizations/Telescoping& Meta-solving)) |
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-Lemonie | -Lemonie | ||
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+ | ==Solution 2== | ||
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+ | If you weren't able to come up with the <math>(3 - 2)</math> insight, then you could just notice that the answer is divisible by | ||
+ | <math>(2 + 3) = 5</math>, and <math>(2^2 + 3^2) = 13</math>. We can then use Fermat's Little Theorem for <math>p = 5, 13</math> on the answer choices to determine which of the answer choices are divisible by both <math>5</math> and <math>13</math>. This is <math>\boxed{C}</math>. | ||
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+ | -MEWTO | ||
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== |
Revision as of 23:25, 11 February 2021
Contents
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
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution by OmegaLearn(Factorizations/Telescoping& Meta-solving)
~ pi_is_3.14
See also
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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.