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

Revision as of 22:47, 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

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 $3^{100}+2^{100} \stackrel{?}{=} 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} \stackrel{?}{=} 3^{100}+3^4\cdot 2^{96}.$ The right side is larger than the left side because $2^{100} \leq 2^{96}\cdot 3^4.$ 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 $\boxed{(A)}$ .

Solution 4

A faster solution. Recognize that for exponents of this size $3^{n}$ will be enormously greater than $2^{n}$ , so the terms involving $2$ will actually have very little effect on the quotient. Now we know the answer will be very close to $81$ .

Notice that the terms being added on to the top and bottom are in the ratio $\frac{1}{16}$ with each other, so they must pull the ratio down from 81 very slightly. (In the same way that a new test score lower than your current cumulative grade always must pull that grade downward.) Answer: $\boxed{(A)}$ .

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

@@ Line 17: / Line 17: @@
 <cmath>3^{100}+2^{100}=x(3^{96}+2^{96}).</cmath>
 Now let's take a look at the answer choices. We notice that <math>81</math>, choice <math>B</math>, can be written as <math>3^4</math>. Plugging this into out equation above, we get
-<cmath>3^{100}+2^{100} \stackrel{?}{=} 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} \stackrel{?}{=} 3^{100}+3^4*2^{96}.</cmath>
+<cmath>3^{100}+2^{100} \stackrel{?}{=} 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} \stackrel{?}{=} 3^{100}+3^4\cdot 2^{96}.</cmath>
 The right side is larger than the left side because
-<cmath>2^{100} \leq 2^{96}*3^4.</cmath>
+<cmath>2^{100} \leq 2^{96}\cdot 3^4.</cmath>
 This means that our original value, <math>x</math>, must be less than <math>81</math>. The only answer that is less than <math>81</math> is <math>80</math> so our answer is <math>\boxed{A}</math>.

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