2021 AMC 12A Problems/Problem 9

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Problem

Which of the following is equivalent to \[(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?\] $\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}$

Solution 1

All you need to do is multiply the entire equation by $(3-2)$. Then all the terms will easily simplify by difference of squares and you will get $3^{128}-2^{128}$ or $\boxed{C}$ as your final answer. Notice you don't need to worry about $3-2$ because that's equal to $1$.

-Lemonie

Solution 2

If you weren't able to come up with the $(3 - 2)$ insight, then you could just notice that the answer is divisible by $(2 + 3) = 5$, and $(2^2 + 3^2) = 13$. We can then use Fermat's Little Theorem for $p = 5, 13$ on the answer choices to determine which of the answer choices are divisible by both $5$ and $13$. This is $\boxed{C}$.

-MEWTO

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

Video Solution by OmegaLearn(Factorizations/Telescoping& Meta-solving)

https://youtu.be/H34IFMlq7Lk

~ pi_is_3.14

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png

2021 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png