# 2021 AMC 12A Problems/Problem 9

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 smugly 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 $x=2$. 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

## Video Solution (HOW TO THINK CREATIVELY!!!)

-Education the Study of Everything

~ pi_is_3.14

~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

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