Difference between revisions of "2013 AMC 10A Problems/Problem 8"
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~Education, the Study of Everything | ~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== | ==Video Solution== | ||
Revision as of 22:54, 2 November 2025
Contents
Problem
What is the value of ![\[\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?\]](http://latex.artofproblemsolving.com/c/c/f/ccfcfc9c986178ae5c0512246b8417f66d28fcfa.png)
Solution
Factoring out, we get: 
.
Cancelling out the 
from the numerator and denominator, we see that it simplifies to 
.
Solution 2
Let 
.
Then the given expression is equal to 
.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Solution 3
We observe that all the answer choices other than 
are integer.
Claim: 
is not an integer.
Proof:
We know that if 
and 
, then 
. So all we need to prove is that 
. Since the numerator is larger than the denominator, the only possible case is 
. 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 
can be factored since is a common factor. Factoring gives us 
. From this we know that 
is a factor of the original expression. Hence 
. So 
. So the fraction cannot be simplified into an integer. Hence the answer is the only fraction, 
.
~JerryZYang
Video Solution
https://www.youtube.com/watch?v=2vf843cvVzo?t=545
~sugar_rush
~savannahsolver
See Also
| 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. 
