2010 AMC 12B Problems/Problem 15
Problem 15
For how many ordered triples 
of nonnegative integers less than 
are there exactly two distinct elements in the set 
, where 
?
Solution
We have either 
, 
, or 
.
For 
, this only occurs at 
. 
has only one solution, namely, 
. 
has five solutions between zero and nineteen, 
, and 
. 
has nineteen integer solutions between zero and nineteen. So for , we have 
ordered triples.
For , again this only occurs at . 
has nineteen solutions, has five solutions, and 
has one solution, so again we have ordered triples.
For , this occurs at and 
. and both have one solution while 
has fifteen solutions. 
and 
both have one solution, namely, 
and , while 
has twenty solutions (
only cycles as 
). So we have 
ordered triples.
In total we have 
ordered triples
~ Small Clarification ~
To more clearly see why the reasoning above is true, try converting the complex numbers into exponential form. That way, we can more easily raise the numbers to 
, 
and 
respectively.
See also
| 2010 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
