Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution)
Line 11: Line 11:
 
==Solution 2 (a variant of Solution 1)==
 
==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 <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>
+
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>
  
−
[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 05:16, 4 January 2023 (EST)
+
~ [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
  
−
=Solution 3 (Lifting the exponent)=
+
==Solution 3 (Lifting the exponent)==
  
 
Let <math>n=13^4-11^4.</math> We wish to find the largest power of <math>2</math> that divides <math>n</math>.
 
Let <math>n=13^4-11^4.</math> We wish to find the largest power of <math>2</math> that divides <math>n</math>.

Revision as of 21:08, 9 August 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

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 $n$,

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

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

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
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_15&oldid=196711"