2018 AMC 10A Problems/Problem 14

Revision as of 01:30, 3 August 2018 by Mathequalsperfect (talk | contribs) (Solution 9)

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

$\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{2^{96}(\frac{3^{100}}{2^{96}})+2^{96}(2^{4})}{2^{96}(\frac{3}{2})^{96}+2^{96}(1)}=\frac{\frac{3^{100}}{2^{96}}+2^{4}}{(\frac{3}{2})^{96}+1}=\frac{\frac{3^{100}}{2^{100}}*2^{4}+2^{4}}{(\frac{3}{2})^{96}+1}=\frac{2^{4}(\frac{3^{100}}{2^{100}}+1)}{(\frac{3}{2})^{96}+1}$.

We can ignore the 1's on the end because they won't really affect the fraction. So, the answer is very very very close but less than the new fraction.

$\frac{2^{4}(\frac{3^{100}}{2^{100}}+1)}{(\frac{3}{2})^{96}+1}<\frac{2^{4}(\frac{3^{100}}{2^{100}})}{(\frac{3}{2})^{96}}$

$\frac{2^{4}(\frac{3^{100}}{2^{100}})}{(\frac{3}{2})^{96}}=\frac{3^{4}}{2^{4}}*2^{4}=3^{4}=81$

So, our final answer is very close but not quite 81, and therefore the greatest integer less than the number is $\boxed{(A) 80}$


Solution 3

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 $\boxed{(A) 80}$ (RegularHexagon)

Solution 4

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 5 (eyeball it)

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 6 (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 7 (The Slick Solution)

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

Solution 8

For positive $a, b, c, d$, if $\frac{a}{b}<\frac{c}{d}$ then $\frac{c+a}{d+b}<\frac{c}{d}$. Let $a=2^{100}, b=2^{96}, c=3^{100}, d=3^{96}$. Then $\frac{c}{d}=3^4$. So answer is less than 81, which leaves only one choice, 80.

  • Note that the algebra here is synonymous to the explanation given in Solution 5. This is the algebraic reason to the logic of if you get a test score with a lower percentage than your average (no matter how many points/percentage of your total grade it was worth), it will pull your overall grade down.

~ ccx09

Solution 9

Try long division, and notice putting $3^4=81$ as denominator is too big and putting $3^4-1=80$ is too small. So we know that the answer is between 80 and 81, yielding 81 as our answer.

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions

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