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

(Solution)
Line 5: Line 5:
 
<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>
  
==Solution==
+
==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>.

Revision as of 18:10, 4 January 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)

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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png