Difference between revisions of "2016 AMC 8 Problems/Problem 15"
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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>. | ||
| − | == Video Solution == | + | == Video Solution by OmegaLearn== |
https://youtu.be/HISL2-N5NVg?t=3705 | https://youtu.be/HISL2-N5NVg?t=3705 | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
| + | == Video Solution== | ||
https://youtu.be/mZCOgH2kVuE | https://youtu.be/mZCOgH2kVuE | ||
Revision as of 17:01, 31 December 2022
Problem
What is the largest power of 
that is a divisor of 
?
Solution
First, we use difference of squares on 
to get 
. Using difference of squares again and simplifying, we get 
. 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 that is a divisor is 
.
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=3705
~ pi_is_3.14
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
