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, we see that the greatest power of <math>2</math> that is a divisor <math>13^4 - 11^4</math> is <math>\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, 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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Revision as of 11:06, 23 November 2016
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, we see that the greatest power of that is a divisor is .
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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All AJHSME/AMC 8 Problems and Solutions |
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