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

Solution 1

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 $\boxed{A}$ .

~Nivek

Solution 2

Let $x=3^{96}$ and $y=2^{96}$ . Then our fraction can be written as $\frac{81x+16y}{x+y}=\frac{16x+16y}{x+y}+\frac{65x}{x+y}=16+\frac{65x}{x+y}$ . Notice that $\frac{65x}{x+y}<\frac{65x}{x}=65$ . So , $16+\frac{65x}{x+y}<16+65=81$ . And our only answer choice less than 81 is $\boxed{(A)}$

~RegularHexagon

Solution 3

Let $x=\frac{3^{100}+2^{100}}{3^{96}+2^{96}}$ . Multiply both sides by $(3^{96}+2^{96})$ , and expand. Rearranging the terms, we get $3^{96}(3^4-x)+2^{96}(2^4-x)=0$ . The left side is strictly decreasing, and it is negative when $x=81$ . This means that the answer must be less than $81$ ; therefore the answer is $\boxed{(A)}$ .

2018 AMC 10A (Problems • Answer Key • Resources)
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2018 AMC 10A Problems/Problem 14

Contents

Solution

Solution 1

Solution 2

Solution 3