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
,
,
Solution 6 (Fast)
Notice that . Hence, . Thus, . Therefore, the answer is
~Mathmagicops
Solution 7
We can rewrite as We do a series of algebraic manipulations:
Let
We let
We notice that which makes look suspiciously like Sure enough, when we try the given condition on we see that it works. We evaluate and get and therefore
~Technodoggo
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.