Difference between revisions of "2016 AMC 8 Problems/Problem 15"

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https://youtu.be/mZCOgH2kVuE
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~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2016|num-b=14|num-a=16}}
 
{{AMC8 box|year=2016|num-b=14|num-a=16}}
 
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Revision as of 14:31, 23 March 2022

Problem

What is the largest power of $2$ that is a divisor of $13^4 - 11^4$?

$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$

Solution

First, we use difference of squares on $13^4 - 11^4 = (13^2)^2 - (11^2)^2$ to get $13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2)$. Using difference of squares again and simplifying, we get $(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)$. Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of $2$ that is a divisor $13^4 - 11^4$ is $\boxed{\textbf{(C)}\ 32}$.

Video Solution

https://youtu.be/HISL2-N5NVg?t=3705

~ pi_is_3.14

https://youtu.be/mZCOgH2kVuE

~savannahsolver

See Also

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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