Difference between revisions of "2016 AMC 10A Problems/Problem 23"
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We see that <math>a \diamond a = 1</math>, and think of division. Testing, we see that the first condition <math>a \diamond (b \diamond c) = (a \diamond b) \cdot c</math> is satisfied, because <math>\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c</math>. Therefore, division is the operation <math>\diamond</math>. Solving the equation, <cmath>\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100\implies x=\frac{100}{336} = \frac{25}{84},</cmath> so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math> | We see that <math>a \diamond a = 1</math>, and think of division. Testing, we see that the first condition <math>a \diamond (b \diamond c) = (a \diamond b) \cdot c</math> is satisfied, because <math>\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c</math>. Therefore, division is the operation <math>\diamond</math>. Solving the equation, <cmath>\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100\implies x=\frac{100}{336} = \frac{25}{84},</cmath> so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math> | ||
− | + | ==Solution 2== | |
− | + | 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>b,</math> <math>( a\ \diamondsuit\ b) = \frac{a}{b}.</math> | |
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+ | 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== | ||
{{AMC10 box|year=2016|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2016|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:33, 5 February 2016
Contents
[hide]Problem
A binary operation has the properties that and that for all nonzero real numbers and . (Here represents multiplication). The solution to the equation can be written as , where and are relatively prime positive integers. What is
Solution
We see that , and think of division. Testing, we see that the first condition is satisfied, because . Therefore, division is the operation . Solving the equation, so the answer is
Solution 2
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 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 10 Problems and Solutions |
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