Difference between revisions of "2016 AMC 10A Problems/Problem 23"
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== Solutions == | == Solutions == | ||
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+ | note that (2016 T 6 )x = 100 where T = the diamond symbol | ||
+ | hence we realies that 2016 T 6 = 2010 T 6 + 6 T 6 = 2004 T 6 + 2*(6 T 6 ) = ... = 336 * ( 6 T 6 ) = 336 | ||
=== Solution 1 === | === Solution 1 === | ||
Revision as of 02:40, 23 December 2020
Contents
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
Solutions
Solution
note that (2016 T 6 )x = 100 where T = the diamond symbol hence we realies that 2016 T 6 = 2010 T 6 + 6 T 6 = 2004 T 6 + 2*(6 T 6 ) = ... = 336 * ( 6 T 6 ) = 336
Solution 1
We see that , and think of division. Testing, we see that the first condition is satisfied, because . Therefore, division can be 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 first 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
Video Solution
https://www.youtube.com/watch?v=8GULAMwu5oE
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.