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

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

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.

@@ Line 9: / Line 9: @@
 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>.
-==Solution 2 (a variant of Solution 1)==
+~CHECKMATE2021
-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)
-=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>.
-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>.
-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(2)+v_2(2)+v_2(24)</cmath>
-<cmath>=1+1+3=5.</cmath>
-Therefore, exponent of the largest power of <math>2</math> that divids <math>13^4-11^4</math> is <math>5,</math> so the largest power of <math>2</math> that divides this number is <math>2^5=\boxed{\textbf{(C)} 32}</math>.
--Benedict T (countmath1)
 == Video Solution by OmegaLearn==
@@ Line 40: / Line 20: @@
 ~savannahsolver
+==Video Solution (CREATIVE THINKING!!!)==
+https://youtu.be/fWEwuLKZ7jY
+~Education, the Study of Everything
 ==See Also==
 {{AMC8 box|year=2016|num-b=14|num-a=16}}
 {{MAA Notice}}

Page

Toolbox

Search