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

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Denote <math>v_p(k)</math> as the largest exponent of <math>p</math> in the prime factorization of <math>n</math>. In this problem, we have <math>p=2</math>.  
 
Denote <math>v_p(k)</math> as the largest exponent of <math>p</math> in the prime factorization of <math>n</math>. In this problem, we have <math>p=2</math>.  
  
−
Lifting the Exponent on <math>n</math>,  
+
By the Lifting the Exponent Lemma on <math>n</math>,  
  
 
<cmath>v_2(13^4-11^4)=v_2(13-11)+v_2(2)+v_2(13+11)</cmath>
 
<cmath>v_2(13^4-11^4)=v_2(13-11)+v_2(2)+v_2(13+11)</cmath>

Revision as of 14:43, 7 March 2023

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

Solution 2 (a variant of Solution 1)

Just like in the above solution, we use the difference-of-squares factorization, but only once to get $13^4-11^4=(13^2-11^2)(13^2+11^2).$ We can then compute that this is equal to $48\cdot290.$ Note that $290=2\cdot145$ (we don't need to factorize any further as $145$ is already odd) thus the largest power of $2$ that divides $290$ is only $2^1=2,$ while $48=2^4\cdot3,$ so the largest power of $2$ that divides $48$ is $2^4=16.$ Hence, the largest power of $2$ that is a divisor of $13^4-11^4$ is $2\cdot16=\boxed{\textbf{(C}~32}.$

Aops-g5-gethsemanea2 (talk) 05:16, 4 January 2023 (EST)

Solution 3 (Lifting the exponent)

Let $n=13^4-11^4.$ We wish to find the largest power of $2$ that divides $n$.

Denote $v_p(k)$ as the largest exponent of $p$ in the prime factorization of $n$. In this problem, we have $p=2$.

By the Lifting the Exponent Lemma on $n$,

\[v_2(13^4-11^4)=v_2(13-11)+v_2(2)+v_2(13+11)\] \[=v_2(2)+v_2(2)+v_2(24)\] \[=1+1+3=5.\]

Therefore, exponent of the largest power of $2$ that divids $13^4-11^4$ is $5,$ so the largest power of $2$ that divides this number is $2^5=\boxed{\textbf{(C)} 32}$.

-Benedict T (countmath1)

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

https://youtu.be/mZCOgH2kVuE

~savannahsolver

See Also

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

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