2016 AMC 10A Problems/Problem 23
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
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 (Proving that is division)
If the given conditions hold for all nonzero numbers and ,
Let From the first two givens, this implies that
From this equation simply becomes
Let Substituting this into the first two conditions, we see that
Substituting , the second equation becomes
Since and are nonzero, we can divide by which yields,
Now we can find the value of straightforwardly:
Therefore,
-Benedict T (countmath1)
Note: We only really cared about what was, so we used the existence of to get an expression in terms of just and .
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
Solution 4
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 5
, , so
,
,
Video Solution 1
https://www.youtube.com/watch?v=8GULAMwu5oE
Video Solution 2(Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=1632
~ pi_is_3.14
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 |
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