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>. | ||
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| + | ==Solution 2 (a variant of Solution 1)== | ||
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| + | 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> | ||
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| + | [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 05:16, 4 January 2023 (EST) | ||
== Video Solution by OmegaLearn== | == Video Solution by OmegaLearn== | ||
Revision as of 06:16, 4 January 2023
Contents
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 
.
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 
We can then compute that this is equal to 
Note that 
(we don't need to factorize any further as 
is already odd) thus the largest power of that divides 
is only 
while 
so the largest power of that divides 
is 
Hence, the largest power of that is a divisor of 
is 
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
~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. 
