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Difference between revisions of "2021 AMC 12A Problems/Problem 9"

(→‎Video Solution by Aaron He)
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== Solution 4 (Engineer's Induction) ==
 
== Solution 4 (Engineer's Induction) ==
 
We can compute some of the first few partial products, and notice that <math>\prod_{k = 0}^{2^n} (2^{2^n}+3^{2^n}) = 3^{2^{n+1}} - 2^{2^{n+1}}</math>. As we don't have to prove this, we get the product is <math>3^{2^7} - 2^{2^7} = 3^{128} - 2^{128}</math>, and smugly click <math>\boxed{\textbf{(C)} ~3^{128} - 2^{128}}</math>. ~rocketsri
 
We can compute some of the first few partial products, and notice that <math>\prod_{k = 0}^{2^n} (2^{2^n}+3^{2^n}) = 3^{2^{n+1}} - 2^{2^{n+1}}</math>. As we don't have to prove this, we get the product is <math>3^{2^7} - 2^{2^7} = 3^{128} - 2^{128}</math>, and smugly click <math>\boxed{\textbf{(C)} ~3^{128} - 2^{128}}</math>. ~rocketsri
 +
 +
== Solution 5(Pure Guessing) ==
 +
Looking at the choices, B and E looks pretty different then the rest of the choice. Remember, MAA won't let you easily get the answer!
 +
Look at A, C, D, two of the solutions include x^128, two solutions are in the form of a - b
 +
Based on MAA's behaviour of letting students getting confused by choices often(which MAA always do that), the solution must include both of the characteristcs(x^128, a-b) or else students can find the correct choice through elimination based on only one characteristics(x^128 or a - b)
 +
 +
Therefore only <math>\boxed{\textbf{(C)} ~3^{128} - 2^{128}}</math> includes the correct solution.
 +
 +
~North America Math Contest Go Go Go
  
 
==Video Solution by Aaron He==
 
==Video Solution by Aaron He==

Revision as of 14:17, 14 February 2021

The following problem is from both the 2021 AMC 10A #10 and 2021 AMC 12A #9, so both problems redirect to this page.

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

Solution 3

After expanding the first few terms, the result after each term appears to be $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$ 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 $2^{2^{n-1}}$ would give $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^{2^{n-1}+1}\cdot{3^{2^{n-1}-1}} + 2^{2^{n-1}}\cdot{3^{2^{n-1}}}$, and all the previous terms multiplied by $3^{2^{n-1}}$ would give $3^{2^n-1} + 3^{2^n-2}\cdot{2^1} + 3^{2^n-3}\cdot{2^2} + ... + 3^{2^{n-1}+1}\cdot{2^{2^{n-1}-1}} + 3^{2^{n-1}}\cdot{2^{2^{n-1}}}$. Their sum is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$, so the proof is complete. Since $\frac{3^{2^n}-2^{2^n}}{3-2}$ is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$, the answer is $\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{C}$.

-SmileKat32

Solution 4 (Engineer's Induction)

We can compute some of the first few partial products, and notice that $\prod_{k = 0}^{2^n} (2^{2^n}+3^{2^n}) = 3^{2^{n+1}} - 2^{2^{n+1}}$. As we don't have to prove this, we get the product is $3^{2^7} - 2^{2^7} = 3^{128} - 2^{128}$, and smugly click $\boxed{\textbf{(C)} ~3^{128} - 2^{128}}$. ~rocketsri

Solution 5(Pure Guessing)

Looking at the choices, B and E looks pretty different then the rest of the choice. Remember, MAA won't let you easily get the answer! Look at A, C, D, two of the solutions include x^128, two solutions are in the form of a - b Based on MAA's behaviour of letting students getting confused by choices often(which MAA always do that), the solution must include both of the characteristcs(x^128, a-b) or else students can find the correct choice through elimination based on only one characteristics(x^128 or a - b)

Therefore only $\boxed{\textbf{(C)} ~3^{128} - 2^{128}}$ includes the correct solution.

~North America Math Contest Go Go Go

Video Solution by Aaron He

https://www.youtube.com/watch?v=xTGDKBthWsw&t=9m30s

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 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

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