Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 4 (Engineer's Induction))
m (→‎Solution 6 (Generalization))
 
(55 intermediate revisions by 15 users not shown)
Line 1: Line 1:
−
{{duplicate|[[2021 AMC 10A Problems#Problem 10|2021 AMC 10A #10]] and [[2021 AMC 12A Problems#Problem 9|2021 AMC 12A #9]]}}
+
{{duplicate|[[2021 AMC 10A Problems/Problem 10|2021 AMC 10A #10]] and [[2021 AMC 12A Problems/Problem 9|2021 AMC 12A #9]]}}
  
 
==Problem==
 
==Problem==
 
Which of the following is equivalent to
 
Which of the following is equivalent to
 
<cmath>(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})?</cmath>
 
<cmath>(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})?</cmath>
−
<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} + 2^{128} \qquad\textbf{(E)} ~5^{127}</math>
  
 
==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}</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>.
+
By multiplying the entire equation by <math>3-2=1</math>, all the terms will simplify by difference of squares, and the final answer is <math>\boxed{\textbf{(C)} ~3^{128}-2^{128}}</math>.
  
−
-Lemonie
+
Additionally, we could also multiply the entire equation (we can let it be equal to <math>x</math>) by <math>2-3=-1</math>. The terms again simplify by difference of squares. This time, we get <math>-x=2^{128}-3^{128} \Rightarrow x=3^{128}-2^{128}</math>. Both solutions yield the same answer.
 +
 
 +
~BakedPotato66
  
 
==Solution 2==
 
==Solution 2==
  
 
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  
 
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>.
+
<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{\textbf{(C)} ~3^{128}-2^{128}}</math>.
  
−
-MEWTO
+
-mewto
  
 
==Solution 3==
 
==Solution 3==
  
−
After expanding the first few terms, the result after each term appears to be <math>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}</math> 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 <math>2^{2^{n-1}}</math> would give <math>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}}}</math>, and all the previous terms multiplied by <math>3^{2^{n-1}}</math> would give <math>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}}}</math>. Their sum is equal to <math>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}</math>, so the proof is complete. Since <math>\frac{3^{2^n}-2^{2^n}}{3-2}</math> is equal to <math>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}</math>, the answer is <math>\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{C}</math>.
+
After expanding the first few terms, the result after each term appears to be <math>2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + \cdots + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}</math> where <math>n</math> 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 <math>2^{2^{n-1}}</math> would give <math>2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + \cdots + 2^{2^{n-1}+1}\cdot{3^{2^{n-1}-1}} + 2^{2^{n-1}}\cdot{3^{2^{n-1}}}</math>, and all the previous terms multiplied by <math>3^{2^{n-1}}</math> would give <math>3^{2^n-1} + 3^{2^n-2}\cdot{2^1} + 3^{2^n-3}\cdot{2^2} + \cdots + 3^{2^{n-1}+1}\cdot{2^{2^{n-1}-1}} + 3^{2^{n-1}}\cdot{2^{2^{n-1}}}</math>. Their sum is equal to <math>2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + \cdots + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}</math>, so the proof is complete. Since <math>\frac{3^{2^n}-2^{2^n}}{3-2}</math> is equal to <math>2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + \cdots + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}</math>, the answer is <math>\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{\textbf{(C)} ~3^{128}-2^{128}}</math>.
  
−
-SmileKat32
+
~SmileKat32
  
 
== 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 click <math>\boxed{\textbf{(C)} ~3^{128}-2^{128}}</math>.  
  
−
== Solution 5(Pure Guessing) ==
+
~rocketsri
−
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.
+
== Solution 5 (Difference of Squares) ==
 +
We notice that the first term is equal to <math>3^2 - 2^2</math>. If we multiply this by the second term, then we will get <math>(3^2 - 2^2)(3^2 + 2^2)</math>, and we can simplify by using difference of squares to obtain <math>3^4 - 2^4</math>. If we multiply this by the third term and simplify using difference of squares again, we get <math>3^8 - 2^8</math>. We can continue down the line until we multiply by the last term, <math>3^{64} + 2^{64}</math>, and get <math>\boxed{\textbf{(C)} ~3^{128}-2^{128}}</math>.
  
−
~North America Math Contest Go Go Go
+
~mathboy100
 +
 
 +
== Solution 6 (Generalization) ==
 +
Let’s generalize to <math>x=3</math> and <math>y=2</math>. Then we get:
 +
<cmath>(y+x)(y^2+x^2)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64}).</cmath>
 +
We see that the first term is <math>y+x</math>, and the next is <math>y^2+x^2</math>. We realize that if we multiply the first term by <math>y-x</math>, the result by difference of squares will be <math>y^2-x^2</math>: we can proceed to use difference of squares on that. In other words, the equation has the domino effect, and all we need to get started is to multiply the whole equation by <math>y-x</math> (keeping in mind to divide <math>y-x</math> at the end.)
 +
 
 +
Now we get:
 +
<cmath>\begin{align*}
 +
&\hspace{5.125mm}(y-x)(y+x)(y^2+x^2)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64}) \\
 +
&=(y^2-x^2)(y^2+x^2)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64}) \\
 +
&=(y^4-x^4)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64}) \\
 +
& \ \vdots \\
 +
&=(y^{64}-x^{64})(y^{64}+x^{64}) \\
 +
&=y^{128}-x^{128}.
 +
\end{align*}</cmath>
 +
Now we can plug in <math>y = 2</math> and <math>x = 3:</math> <cmath>(2-3)(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}) = 2^{128}-3^{128}.</cmath>
 +
However, we must not forget to divide by <math>2-3 = -1</math> at the end! Dividing, we get the answer of
 +
<cmath>3^{128}-2^{128} = \boxed{\textbf{(C)} ~3^{128}-2^{128}}.</cmath>
 +
~KingRavi
 +
 
 +
== Solution 7 (Generalization) ==
 +
More generally, we have <cmath>\bigl(a^n-b^n\bigr)\bigl(a^n+b^n\bigr)=a^{2n}-b^{2n}\hspace{20mm}(\bigstar)</cmath> by the difference of squares.
 +
 
 +
Multiplying the original expression by <math>1=3-2</math> and then applying <math>(\bigstar)</math> repeatedly, we get
 +
<cmath>\begin{align*}
 +
&\hspace{5.125mm}\bigl(2+3\bigr)\bigl(2^2+3^2\bigr)\bigl(2^4+3^4\bigr)\bigl(2^8+3^8\bigr)\bigl(2^{16}+3^{16}\bigr)\bigl(2^{32}+3^{32}\bigr)\bigl(2^{64}+3^{64}\bigr) \\
 +
&=\underbrace{\bigl(\hspace{3.25mm}1\hspace{3.25mm}\bigr)}_{3-2}\underbrace{\bigl(2+3\bigr)\bigl(2^2+3^2\bigr)\bigl(2^4+3^4\bigr)\bigl(2^8+3^8\bigr)\bigl(2^{16}+3^{16}\bigr)\bigl(2^{32}+3^{32}\bigr)\bigl(2^{64}+3^{64}\bigr)}_{\text{Apply the }\textbf{Commutative Property of Addition}\text{ to all factors.}} \\
 +
&=\underbrace{\bigl(3-2\bigr)\bigl(3+2\bigr)}_{\substack{\text{Apply }(\bigstar) \\ \text{to get }3^2-2^2.}}\bigl(3^2+2^2\bigr)\bigl(3^4+2^4\bigr)\bigl(3^8+2^8\bigr)\bigl(3^{16}+2^{16}\bigr)\bigl(3^{32}+2^{32}\bigr)\bigl(3^{64}+2^{64}\bigr) \\
 +
&=\underbrace{\bigl(3^2-2^2\bigr)\bigl(3^2+2^2\bigr)}_{\substack{\text{Apply }(\bigstar) \\ \text{to get }3^4-2^4.}}\bigl(3^4+2^4\bigr)\bigl(3^8+2^8\bigr)\bigl(3^{16}+2^{16}\bigr)\bigl(3^{32}+2^{32}\bigr)\bigl(3^{64}+2^{64}\bigr) \\
 +
&=\underbrace{\bigl(3^4-2^4\bigr)\bigl(3^4+2^4\bigr)}_{\substack{\text{Apply }(\bigstar) \\ \text{to get }3^8-2^8.}}\bigl(3^8+2^8\bigr)\bigl(3^{16}+2^{16}\bigr)\bigl(3^{32}+2^{32}\bigr)\bigl(3^{64}+2^{64}\bigr) \\
 +
& \ \vdots \\
 +
&=\boxed{\textbf{(C)} ~3^{128}-2^{128}}.
 +
\end{align*}</cmath>
 +
~MRENTHUSIASM
 +
 
 +
==Solution 8 (Desperate & Cheese, Similar to Solution 4)==
 +
One would probably assume that the factorization that this is equal to can be generalized for the pattern <cmath>(2+3)(2^2 + 3^2)(2^4+3^4)\cdots(2^n + 3^n).</cmath>
 +
So we can take a smaller portion and apply the answer choices there. We get
 +
<cmath>(2+3)(2^2 + 3^2) = (5)(13) = 65.</cmath>
 +
We notice that the patterns have a relationship where its either the sum of the powers or the next power not included that is incorporated. So we change the numbers from <math>127</math> to <math>3</math>, <math>128</math> to <math>4</math>, and <math>63</math> to <math>1</math> (since <math>63</math> is the sum of the powers, divided by <math>2</math>, rounded down). Checking the values of the other patterns, the only one that gives <math>65</math> is <math>3^4 - 2^4</math>, which means the answer is <math>\boxed{\textbf{(C)} ~3^{128}-2^{128}}.</math>
 +
 
 +
~neeyakkid23
 +
 
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/Pm3euI3jyDk
 +
 
 +
-Education the Study of Everything
  
 
==Video Solution by Aaron He==
 
==Video Solution by Aaron He==
 
https://www.youtube.com/watch?v=xTGDKBthWsw&t=9m30s
 
https://www.youtube.com/watch?v=xTGDKBthWsw&t=9m30s
 +
 +
==Video Solution (Conjugation, Difference of Squares)==
 +
https://www.youtube.com/watch?v=gXaIyeMF7Qo&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=9
 +
 
==Video Solution by Hawk Math==
 
==Video Solution by Hawk Math==
 
https://www.youtube.com/watch?v=P5al76DxyHY
 
https://www.youtube.com/watch?v=P5al76DxyHY
  
−
== Video Solution by OmegaLearn(Factorizations/Telescoping& Meta-solving) ==
+
== Video Solution by OmegaLearn (Factorizations/Telescoping & Meta-solving) ==
 
https://youtu.be/H34IFMlq7Lk
 
https://youtu.be/H34IFMlq7Lk
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/-MJXKZowfO0
 +
 +
~savannahsolver
 +
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/s6E4E06XhPU?t=771 (for AMC 10A)
 +
 +
https://youtu.be/cckGBU2x1zg?t=548 (for AMC 12A)
 +
 +
~IceMatrix
 +
 +
==Video Solution by The Learning Royal==
 +
https://youtu.be/AWjOeBFyeb4
  
 
==See also==
 
==See also==

Latest revision as of 02:09, 31 October 2024

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

Solution 1

By multiplying the entire equation by $3-2=1$, all the terms will simplify by difference of squares, and the final answer is $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$.

Additionally, we could also multiply the entire equation (we can let it be equal to $x$) by $2-3=-1$. The terms again simplify by difference of squares. This time, we get $-x=2^{128}-3^{128} \Rightarrow x=3^{128}-2^{128}$. Both solutions yield the same answer.

~BakedPotato66

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 $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$.

-mewto

Solution 3

After expanding the first few terms, the result after each term appears to be $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + \cdots + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$ 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 $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + \cdots + 2^{2^{n-1}+1}\cdot{3^{2^{n-1}-1}} + 2^{2^{n-1}}\cdot{3^{2^{n-1}}}$, and all the previous terms multiplied by $3^{2^{n-1}}$ would give $3^{2^n-1} + 3^{2^n-2}\cdot{2^1} + 3^{2^n-3}\cdot{2^2} + \cdots + 3^{2^{n-1}+1}\cdot{2^{2^{n-1}-1}} + 3^{2^{n-1}}\cdot{2^{2^{n-1}}}$. Their sum is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + \cdots + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$, so the proof is complete. Since $\frac{3^{2^n}-2^{2^n}}{3-2}$ is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + \cdots + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$, the answer is $\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{\textbf{(C)} ~3^{128}-2^{128}}$.

~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 click $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$.

~rocketsri

Solution 5 (Difference of Squares)

We notice that the first term is equal to $3^2 - 2^2$. If we multiply this by the second term, then we will get $(3^2 - 2^2)(3^2 + 2^2)$, and we can simplify by using difference of squares to obtain $3^4 - 2^4$. If we multiply this by the third term and simplify using difference of squares again, we get $3^8 - 2^8$. We can continue down the line until we multiply by the last term, $3^{64} + 2^{64}$, and get $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$.

~mathboy100

Solution 6 (Generalization)

Let’s generalize to $x=3$ and $y=2$. Then we get: \[(y+x)(y^2+x^2)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64}).\] We see that the first term is $y+x$, and the next is $y^2+x^2$. We realize that if we multiply the first term by $y-x$, the result by difference of squares will be $y^2-x^2$: we can proceed to use difference of squares on that. In other words, the equation has the domino effect, and all we need to get started is to multiply the whole equation by $y-x$ (keeping in mind to divide $y-x$ at the end.)

Now we get: \begin{align*} &\hspace{5.125mm}(y-x)(y+x)(y^2+x^2)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64}) \\ &=(y^2-x^2)(y^2+x^2)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64}) \\ &=(y^4-x^4)(y^4+x^4)(y^8+x^8)(y^{16}+x^{16})(y^{32}+x^{32})(y^{64}+x^{64}) \\ & \ \vdots \\ &=(y^{64}-x^{64})(y^{64}+x^{64}) \\ &=y^{128}-x^{128}. \end{align*} Now we can plug in $y = 2$ and $x = 3:$ \[(2-3)(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64}) = 2^{128}-3^{128}.\] However, we must not forget to divide by $2-3 = -1$ at the end! Dividing, we get the answer of \[3^{128}-2^{128} = \boxed{\textbf{(C)} ~3^{128}-2^{128}}.\] ~KingRavi

Solution 7 (Generalization)

More generally, we have \[\bigl(a^n-b^n\bigr)\bigl(a^n+b^n\bigr)=a^{2n}-b^{2n}\hspace{20mm}(\bigstar)\] by the difference of squares.

Multiplying the original expression by $1=3-2$ and then applying $(\bigstar)$ repeatedly, we get \begin{align*} &\hspace{5.125mm}\bigl(2+3\bigr)\bigl(2^2+3^2\bigr)\bigl(2^4+3^4\bigr)\bigl(2^8+3^8\bigr)\bigl(2^{16}+3^{16}\bigr)\bigl(2^{32}+3^{32}\bigr)\bigl(2^{64}+3^{64}\bigr) \\ &=\underbrace{\bigl(\hspace{3.25mm}1\hspace{3.25mm}\bigr)}_{3-2}\underbrace{\bigl(2+3\bigr)\bigl(2^2+3^2\bigr)\bigl(2^4+3^4\bigr)\bigl(2^8+3^8\bigr)\bigl(2^{16}+3^{16}\bigr)\bigl(2^{32}+3^{32}\bigr)\bigl(2^{64}+3^{64}\bigr)}_{\text{Apply the }\textbf{Commutative Property of Addition}\text{ to all factors.}} \\ &=\underbrace{\bigl(3-2\bigr)\bigl(3+2\bigr)}_{\substack{\text{Apply }(\bigstar) \\ \text{to get }3^2-2^2.}}\bigl(3^2+2^2\bigr)\bigl(3^4+2^4\bigr)\bigl(3^8+2^8\bigr)\bigl(3^{16}+2^{16}\bigr)\bigl(3^{32}+2^{32}\bigr)\bigl(3^{64}+2^{64}\bigr) \\ &=\underbrace{\bigl(3^2-2^2\bigr)\bigl(3^2+2^2\bigr)}_{\substack{\text{Apply }(\bigstar) \\ \text{to get }3^4-2^4.}}\bigl(3^4+2^4\bigr)\bigl(3^8+2^8\bigr)\bigl(3^{16}+2^{16}\bigr)\bigl(3^{32}+2^{32}\bigr)\bigl(3^{64}+2^{64}\bigr) \\ &=\underbrace{\bigl(3^4-2^4\bigr)\bigl(3^4+2^4\bigr)}_{\substack{\text{Apply }(\bigstar) \\ \text{to get }3^8-2^8.}}\bigl(3^8+2^8\bigr)\bigl(3^{16}+2^{16}\bigr)\bigl(3^{32}+2^{32}\bigr)\bigl(3^{64}+2^{64}\bigr) \\ & \ \vdots \\ &=\boxed{\textbf{(C)} ~3^{128}-2^{128}}. \end{align*} ~MRENTHUSIASM

Solution 8 (Desperate & Cheese, Similar to Solution 4)

One would probably assume that the factorization that this is equal to can be generalized for the pattern \[(2+3)(2^2 + 3^2)(2^4+3^4)\cdots(2^n + 3^n).\] So we can take a smaller portion and apply the answer choices there. We get \[(2+3)(2^2 + 3^2) = (5)(13) = 65.\] We notice that the patterns have a relationship where its either the sum of the powers or the next power not included that is incorporated. So we change the numbers from $127$ to $3$, $128$ to $4$, and $63$ to $1$ (since $63$ is the sum of the powers, divided by $2$, rounded down). Checking the values of the other patterns, the only one that gives $65$ is $3^4 - 2^4$, which means the answer is $\boxed{\textbf{(C)} ~3^{128}-2^{128}}.$

~neeyakkid23

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/Pm3euI3jyDk

-Education the Study of Everything

Video Solution by Aaron He

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

Video Solution (Conjugation, Difference of Squares)

https://www.youtube.com/watch?v=gXaIyeMF7Qo&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=9

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

Video Solution by WhyMath

https://youtu.be/-MJXKZowfO0

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/s6E4E06XhPU?t=771 (for AMC 10A)

https://youtu.be/cckGBU2x1zg?t=548 (for AMC 12A)

~IceMatrix

Video Solution by The Learning Royal

https://youtu.be/AWjOeBFyeb4

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_9&oldid=231056"