Difference between revisions of "2016 AMC 10A Problems/Problem 23"
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==Solution== | ==Solution== | ||
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===Solution 1=== | ===Solution 1=== | ||
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===Solution 2=== | ===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 < | + | 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 <cmath>( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\ b) = a\ \diamondsuit\ 1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a.</cmath> 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> |
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> | 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> | ||
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One way to eliminate the <math>\diamondsuit</math> in this equation is to make <math>a = b</math> so that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = c</math>. In this case, we can make <math>b = 2016</math>. | One way to eliminate the <math>\diamondsuit</math> in this equation is to make <math>a = b</math> so that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = c</math>. In this case, we can make <math>b = 2016</math>. | ||
− | < | + | <cmath>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100, |
− | (2016\, \diamondsuit\, 6) \cdot x = 100</ | + | (2016\, \diamondsuit\, 6) \cdot x = 100</cmath> |
By multiplying both sides by <math>\frac{6}{x}</math>, we get: | By multiplying both sides by <math>\frac{6}{x}</math>, we get: | ||
− | < | + | <cmath>(2016\, \diamondsuit\, 6) \cdot 6 = \frac{600}{x}, |
− | 2016 \, \diamondsuit\, (6\, \diamondsuit\, 6) = \frac{600}{x}</ | + | 2016 \, \diamondsuit\, (6\, \diamondsuit\, 6) = \frac{600}{x}</cmath> |
− | Because <math>6\, \diamondsuit\, 6 = 2016\, \diamondsuit\, 2016 = 1</math> | + | Because <math>6\, \diamondsuit\, 6 = 2016\, \diamondsuit\, 2016 = 1:</math> |
− | < | + | <cmath>2016 \, \diamondsuit\, (2016\, \diamondsuit\, 2016) = \frac{600}{x}, |
− | (2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x} | + | (2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x}, |
− | 2016 = \frac{600}{x}</ | + | 2016 = \frac{600}{x}</cmath> |
Therefore, <math>x = \frac{600}{2016} = \frac{25}{84}</math>, so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math> | Therefore, <math>x = \frac{600}{2016} = \frac{25}{84}</math>, so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math> |
Revision as of 17:48, 7 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
Solution 1
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
Solution 3
One way to eliminate the in this equation is to make so that . In this case, we can make .
By multiplying both sides by , we get:
Because
Therefore, , 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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.