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 | |
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