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Difference between revisions of "2016 AMC 8 Problems/Problem 15"
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<math>\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128</math>
 
<math>\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128</math>
  
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==Solution==
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==Solution 1==
  
 
First, we use difference of squares on <math>13^4 - 11^4 = (13^2)^2 - (11^2)^2 </math> to get <math> 13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2) </math>. Using difference of squares again and simplifying, we get <math>(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)</math>. 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 <math>2</math> that is a divisor <math>13^4 - 11^4</math> is <math>\boxed{\textbf{(C)}\ 32}</math>.
 
First, we use difference of squares on <math>13^4 - 11^4 = (13^2)^2 - (11^2)^2 </math> to get <math> 13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2) </math>. Using difference of squares again and simplifying, we get <math>(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)</math>. 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 <math>2</math> that is a divisor <math>13^4 - 11^4</math> is <math>\boxed{\textbf{(C)}\ 32}</math>.
  
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==Solution 2 (a variant of Solution 1)==
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~CHECKMATE2021
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Just like in the above solution, we use the difference-of-squares factorization, but only once to get <math>13^4-11^4=(13^2-11^2)(13^2+11^2).</math> We can then compute that this is equal to <math>48\cdot290.</math> Note that <math>290=2\cdot145</math> (we don't need to factorize any further as <math>145</math> is already odd) thus the largest power of <math>2</math> that divides <math>290</math> is only <math>2^1=2,</math> while <math>48=2^4\cdot3,</math> so the largest power of <math>2</math> that divides <math>48</math> is <math>2^4=16.</math> Hence, the largest power of <math>2</math> that is a divisor of <math>13^4-11^4</math> is <math>2\cdot16=\boxed{\textbf{(C}~32}.</math>
 
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[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 05:16, 4 January 2023 (EST)
 
  
 
== Video Solution by OmegaLearn==
 
== Video Solution by OmegaLearn==
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~savannahsolver
 
~savannahsolver
 +
 +
==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/fWEwuLKZ7jY
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 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2016|num-b=14|num-a=16}}
 
{{AMC8 box|year=2016|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:58, 17 May 2024

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 1

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 $\boxed{\textbf{(C)}\ 32}$.

~CHECKMATE2021

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

https://youtu.be/mZCOgH2kVuE

~savannahsolver

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/fWEwuLKZ7jY

~Education, the Study of Everything

See Also

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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