Difference between revisions of "2010 AMC 12B Problems/Problem 15"
m (→Solution) |
Chezpotato (talk | contribs) (→Solution) |
||
Line 14: | Line 14: | ||
In total we have <math>{95+95+35=\boxed{\text{(D) }225}}</math> ordered triples | In total we have <math>{95+95+35=\boxed{\text{(D) }225}}</math> 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 <math>x</math>, <math>y</math> and <math>z</math> respectively. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=14|num-a=16|ab=B}} | {{AMC12 box|year=2010|num-b=14|num-a=16|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:06, 18 October 2020
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.