Difference between revisions of "2016 AMC 12A Problems/Problem 20"
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We can manipulate the given identities to arrive at a conclusion about the binary operator <math>\diamondsuit</math>. Substituting <math>b = c</math> into the second identity yields <math>( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\ b) = a\ \diamondsuit\ 1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a</math>. Hence, <math>( a\ \diamondsuit\ b) \cdot b = a,</math> or, dividing both sides of the equation by <math>a,</math> <math>( a\ \diamondsuit\ b) = \frac{a}{b}.</math> | We can manipulate the given identities to arrive at a conclusion about the binary operator <math>\diamondsuit</math>. Substituting <math>b = c</math> into the second identity yields <math>( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\ b) = a\ \diamondsuit\ 1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a</math>. Hence, <math>( a\ \diamondsuit\ b) \cdot b = a,</math> or, dividing both sides of the equation by <math>a,</math> <math>( a\ \diamondsuit\ b) = \frac{a}{b}.</math> | ||
− | Hence, the given equation becomes <math>\frac{2016}{\frac{ | + | Hence, the given equation becomes <math>\frac{2016}{\frac{6}{x}} = 100</math>. Solving yields <math>x=\frac{100}{336} = \frac{25}{84},</math> so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math> |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2016|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:03, 4 February 2016
Problem
A binary operation has the properties that and that for all nonzero real numbers and (Here the dot represents the usual multiplication operation.) The solution to the equation can be written as where and are relativelt prime positive integers. What is
Solution
We can manipulate the given identities to arrive at a conclusion about the binary operator . Substituting into the second identity yields . Hence, or, dividing both sides of the equation by
Hence, the given equation becomes . Solving yields so the answer is
See Also
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 AMC 12 Problems and Solutions |
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