Difference between revisions of "2021 AMC 12A Problems/Problem 9"
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==Solution 1== | ==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>. | + | 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}</math> or <math>\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>. |
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| + | -Lemonie | ||
==Note== | ==Note== | ||
Revision as of 15:11, 11 February 2021
Contents
Problem
Which of the following is equivalent to
![\[(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?\]](http://latex.artofproblemsolving.com/6/5/a/65a7155ec29794e20977d4bf1eef21987af9bbee.png)
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
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
