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

(Solution 3 (Lifting the exponent))
(Solution 4 (Brute Force))
 
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~CHECKMATE2021
 
~CHECKMATE2021
 
== Solution 4 (Brute Force) ==
 
We can simply take 13 to the 4th power, which is 28561. We subtract that by 11 to the 4th power, which is 14641 (You can use Pascal's Triangle to find this). Finally, subtract the numbers to get 13920.
 
 
To test the options, since we need the largest one, we can go from top down. Testing, we see that both D and E are decimals,
 
and 32 works. So, our answer is <math>\boxed{\textbf{(C)}~32}.</math>
 
 
-themathgood
 
 
 
 
Note: This is not the fastest way, and is not recommended.
 
 
 
I completely agree even though I did it./reader
 
  
 
== Video Solution by OmegaLearn==
 
== Video Solution by OmegaLearn==

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 (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

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