2018 AMC 10A Problems/Problem 14

Revision as of 15:43, 8 February 2018 by Nivek (talk | contribs) (Solution)

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 $\boxed{A}$.

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