Difference between revisions of "2016 AMC 10A Problems/Problem 23"
m (minor edit) |
(→Solutions) |
||
Line 5: | Line 5: | ||
== Solutions == | == 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 === | === 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.