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Difference between revisions of "2018 AMC 10A Problems/Problem 14"

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~RegularHexagon
 
~RegularHexagon
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==Solution 3==
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Let <math>x=\frac{3^{100}+2^{100}}{3^{96}+2^{96}}</math>. Multiply both sides by <math>(3^{96}+2^{96})</math>, and expand. Rearranging the terms, we get <math>3^{96}(3^4-x)+2^{96}(2^4-x)=0</math>. The left side is strictly decreasing, and it is negative when <math>x=81</math>. Therefore the answer must be less than <math>81</math>, therefore the answer is <math>\boxed{(A)}</math>.
  
 
{{AMC10 box|year=2018|ab=A|num-b=13|num-a=15}}
 
{{AMC10 box|year=2018|ab=A|num-b=13|num-a=15}}
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==Solution 3==
 
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Let <math>x=\frac{3^{100}+2^{100}}{3^{96}+2^{96}}</math>. Multiply both sides by <math>(3^{96}+2^{96})</math>, and expand. Rearranging the terms, we get <math>3^{96}(3^4-x)+2^{96}(2^4-x)=0</math>. The left side is strictly decreasing, and it is negative when <math>x=81</math>. Therefore the answer must be less than <math>81</math>, therefore the answer is <math>\boxed{(A)}</math>.
 

Revision as of 17:47, 8 February 2018

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 $\boxed{A}$.

~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 $x=81$. Therefore the answer must be less than $81$, therefore the answer is $\boxed{(A)}$.

2018 AMC 10A (Problems • Answer Key • Resources)
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Problem 13 		Followed by
Problem 15
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All AMC 10 Problems and Solutions
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