Difference between revisions of "2016 AMC 10A Problems/Problem 23"
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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 | + | <cmath>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100\implies |
(2016\, \diamondsuit\, 6) \cdot x = 100</cmath> | (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} | + | <cmath>(2016\, \diamondsuit\, 6) \cdot 6 = \frac{600}{x}\implies |
2016 \, \diamondsuit\, (6\, \diamondsuit\, 6) = \frac{600}{x}</cmath> | 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} | + | <cmath>2016 \, \diamondsuit\, (2016\, \diamondsuit\, 2016) = \frac{600}{x}\implies |
− | (2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x} | + | (2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x}\implies |
2016 = \frac{600}{x}</cmath> | 2016 = \frac{600}{x}</cmath> | ||
Revision as of 22:53, 8 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.