Difference between revisions of "2016 AMC 10A Problems/Problem 23"
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==Solution== | ==Solution== | ||
− | 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, < | + | 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> |
==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 08:18, 4 February 2016
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
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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