2018 AMC 10A Problems/Problem 14

Revision as of 22:35, 9 February 2018 by Harsha12345 (talk | contribs) (Solution 6)

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 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 $\boxed{A}$.


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 $\boxed{(A)}$


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: $\boxed{(A)}$.

Solution 5 (Using the answer choices)

We can compare the given value to each of our answer choices. We already know that it is greater than $80$ because otherwise there would have been a smaller answer, so we move onto $81$. We get:

$\frac{3^{100}+2^{100}}{3^{96}+2^{96}} \text{ ? } 3^4$

Cross multiply to get:

$3^{100}+2^{100} \text{ ? }3^{100}+(2^{96})(3^4)$

Cancel out $3^{100}$ and divide by $2^{96}$ to get $2^{4} \text{ ? }3^4$. We know that $2^4 < 3^4$, which means the expression is less than $81$ so the answer is $\boxed{(A)}$.

Solution 6(logic)

Notice how $\frac{3^{100}+2^{100}}{3^{96}+2^{96}}$ can be rewritten as $\frac{81(3^{96})+16(2^{96})}{3^{96}+2^{96}}=\frac{81(3^{96})+81(2^{96})}{3^{96}+2^{96}}-\frac{65(2^{96})}{3^{96}+2^{96}}=81-\frac{65(2^{96})}{3^{96}+2^{96}}$. Note that $\frac{65(2^{96})}{3^{96}+2^{96}}<1$, so the greatest integer less than or equal to $\frac{3^{100}+2^{100}}{3^{96}+2^{96}}$ is $80$ or $\boxed{\textbf{(A)}}$ ~blitzkrieg21

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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