Difference between revisions of "2016 AMC 8 Problems/Problem 15"
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==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: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 20:08, 9 August 2023
Contents
[hide]Problem
What is the largest power of that is a divisor of ?
Solution 1
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
Solution 3 (Lifting the exponent)
Let We wish to find the largest power of that divides .
Denote as the largest exponent of in the prime factorization of . In this problem, we have .
By the Lifting the Exponent Lemma on ,
Therefore, exponent of the largest power of that divids is so the largest power of that divides this number is .
-Benedict T (countmath1)
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=3705
~ pi_is_3.14
Video Solution
~savannahsolver
Video Solution (CREATIVE THINKING!!!)
~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.