Difference between revisions of "2021 AMC 12A Problems/Problem 9"
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<math>\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}</math> | <math>\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}</math> | ||
− | ==Solution== | + | ==Solution 1== |
+ | |||
+ | All you need to do is multiply the entire equation by <math>(3-2)</math>. Then all the terms will easily simplify by difference of squares and you will get <math>3^{128}-2^{128} or \boxed{C}</math> as your final answer. Notice you don't need to worry about <math>3-2</math> because that's equal to <math>1</math>. | ||
==Note== | ==Note== |
Revision as of 14:11, 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 as your final answer. Notice you don't need to worry about because that's equal to .
Note
See problem 1.
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 |
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