Difference between revisions of "2010 AMC 12B Problems/Problem 15"

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(Solution)
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== Solution ==
 
== Solution ==
We have either <math>{i^{x}=(1+i)^{y}=/=z}</math>, <math>{i^{x}=z=/=(1+i)^{y}}</math>, or <math>{(1+i)^{y}=z=/=i^x}</math>.
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We have either <math>{i^{x}=(1+i)^{y}neqz}</math>, <math>{i^{x}=zneq(1+i)^{y}}</math>, or <math>{(1+i)^{y}=zneqi^x}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}
 
{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}

Revision as of 14:10, 7 November 2010

Problem 15

For how many ordered triples $(x,y,z)$ of nonnegative integers less than $20$ are there exactly two distinct elements in the set $\{i^x, (1+i)^y, z\}$, where $i=\sqrt{-1}$?

$\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235$

Solution

We have either ${i^{x}=(1+i)^{y}neqz}$, ${i^{x}=zneq(1+i)^{y}}$, or ${(1+i)^{y}=zneqi^x}$.

See also

2010 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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