# 2018 AMC 10A Problems/Problem 14

What is the greatest integer less than or equal to $$\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?$$

$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$

## Solution

Let's set this value equal to $x$. We can write $$\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=x.$$ Multiplying by $3^{96}+2^{96}$ on both sides, we get $$3^{100}+2^{100}=x(3^{96}+2^{96}).$$ Now let's take a look at the answer choices. We notice that $81$, choice $B$, can be written as 3^4. Plugging this into out equation above, we get $$3^{100}+2^{100} \stackrel{?}{=} 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} \stackrel{?}{=} 3^{100}+3^4*2^{96}.$$ The right side is larger than the left side because $$2^{100} \leq 2^{96}*3^4.$$ This means that our original value, $x$, must be less than $81$. The only answer that is less than $81$ is $80$ so our answer is $\boxed{A}$.