Difference between revisions of "2013 AMC 10A Problems/Problem 8"

Latest revision as of 22:56, 2 November 2025

Problem

What is the value of $\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?$

$\textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \frac{5}{3} \qquad\textbf{(D)}\ 2013 \qquad\textbf{(E)}\ 2^{4024}$

Solution

Factoring out, we get: $\frac{2^{2012}(2^2 + 1)}{2^{2012}(2^2-1)}$ .

Cancelling out the $2^{2012}$ from the numerator and denominator, we see that it simplifies to $\boxed{\textbf{(C) }\frac{5}{3}}$ .

Solution 2

Let $x=2^{2012}$ .

Then the given expression is equal to $\frac{4x+x}{4x-x}=\frac{5x}{3x}=\boxed{\textbf{(C) }\frac{5}{3}}$ .

Video Solution (CREATIVE THINKING)

https://youtu.be/1ULw0x9XK4o

~Education, the Study of Everything

Solution 3

We observe that all the answer choices other than $\textbf{(C)}$ are integer.

Claim: $\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}$ is not an integer.

Proof: We know that if $b>a$ and $\operatorname{gcd}(a,b) = a$ , then $a|b$ .

So all we need to prove is that $\operatorname{gcd}(2^{2014}+2^{2012}, 2^{2014}-2^{2012}) \neq 2^{2014}+2^{2012} \text{ or } 2^{2014}-2^{2012}$ . Since the numerator is larger than the denominator, the only possible case is $\gcd(2^{2014}+2^{2012}, 2^{2014}-2^{2012}) = 2^{2014}-2^{2012}$ . We have to prove that this is impossible. Using the Euclidean Algorithm we can simplify to obtain \begin{align*} \gcd(2^{2014}+2^{2012}, 2^{2014}-2^{2012}) &= \gcd(2^{2014}+2^{2012}-2^{2014}+2^{2012}, 2^{2014}-2^{2012})\\ &= \gcd(2 \cdot 2^{2012}, 2^{2014}-2^{2012})\\ &= \gcd(2^{2013}, 2^{2014}-2^{2012}) \end{align*} We notice that $2^{2014}-2^{2012}$ can be factored since $2^{2012}$ is a common factor. Factoring gives us $2^{2014}-2^{2012} = 2^{2012}(2^2-1) = 3 \cdot 2^{2012}$ . From this we know that $3$ is a factor of the original expression. Hence $\gcd(2^{2013}, 2^{2014}-2^{2012}) = \gcd(2^{2013}, 3 \cdot 2^{2012}) = 2^{2013} \neq 2^{2012} \cdot 3$ . So $2^{2014}-2^{2012} \nmid 2^{2014}+2^{2012}$ . So the fraction cannot be simplified into an integer. Hence the answer is the only fraction, $\boxed{\textbf{(C)}}$ .

~JerryZYang

Video Solution

https://www.youtube.com/watch?v=2vf843cvVzo?t=545

~sugar_rush

https://youtu.be/C3prgokOdHc

~savannahsolver

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly  ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
+==Problem==
+What is the value of <cmath>\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?</cmath>
-<math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25 </math>
+<math> \textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1  \qquad\textbf{(C)}\ \frac{5}{3} \qquad\textbf{(D)}\ 2013 \qquad\textbf{(E)}\ 2^{4024} </math>
+==Solution==
+Factoring out, we get: <math>\frac{2^{2012}(2^2 + 1)}{2^{2012}(2^2-1)}</math>.
+Cancelling out the <math>2^{2012}</math> from the numerator and denominator, we see that it simplifies to <math>\boxed{\textbf{(C) }\frac{5}{3}}</math>.
+==Solution 2==
+Let <math>x=2^{2012}</math>.
+Then the given expression is equal to <math>\frac{4x+x}{4x-x}=\frac{5x}{3x}=\boxed{\textbf{(C) }\frac{5}{3}}</math>.
+==Video Solution (CREATIVE THINKING)==
+https://youtu.be/1ULw0x9XK4o
+~Education, the Study of Everything
+==Solution 3==
+We observe that all the answer choices other than <math>\textbf{(C)}</math> are integer.
+Claim: <math>\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}</math> is not an integer.
+Proof:
+We know that if <math>b>a</math> and <math>\operatorname{gcd}(a,b) = a</math>, then <math>a|b</math>.
+So all we need to prove is that <math>\operatorname{gcd}(2^{2014}+2^{2012}, 2^{2014}-2^{2012}) \neq 2^{2014}+2^{2012} \text{ or } 2^{2014}-2^{2012}</math>. Since the numerator is larger than the denominator, the only possible case is <math>\gcd(2^{2014}+2^{2012}, 2^{2014}-2^{2012}) = 2^{2014}-2^{2012}</math>. We have to prove that this is impossible. Using the Euclidean Algorithm we can simplify to obtain
+\begin{align*}
+\gcd(2^{2014}+2^{2012}, 2^{2014}-2^{2012}) &= \gcd(2^{2014}+2^{2012}-2^{2014}+2^{2012}, 2^{2014}-2^{2012})\\
+&= \gcd(2 \cdot 2^{2012}, 2^{2014}-2^{2012})\\
+&= \gcd(2^{2013}, 2^{2014}-2^{2012})
+\end{align*}
+We notice that <math>2^{2014}-2^{2012}</math> can be factored since <math>2^{2012}</math> is a common factor. Factoring gives us <math>2^{2014}-2^{2012} = 2^{2012}(2^2-1) = 3 \cdot 2^{2012}</math>. From this we know that <math>3</math> is a factor of the original expression. Hence <math>\gcd(2^{2013}, 2^{2014}-2^{2012}) = \gcd(2^{2013}, 3 \cdot 2^{2012}) = 2^{2013} \neq 2^{2012} \cdot 3</math>. So <math>2^{2014}-2^{2012} \nmid 2^{2014}+2^{2012}</math>. So the fraction cannot be simplified into an integer. Hence the answer is the only fraction, <math>\boxed{\textbf{(C)}}</math>.
+~JerryZYang
+==Video Solution==
+https://www.youtube.com/watch?v=2vf843cvVzo?t=545
+~sugar_rush
+https://youtu.be/C3prgokOdHc
+~savannahsolver
+==See Also==
+{{AMC10 box|year=2013|ab=A|num-b=7|num-a=9}}
+{{AMC12 box|year=2013|ab=A|num-b=3|num-a=5}}
+{{MAA Notice}}

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