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

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== Problem 15 ==
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== Problem ==
 
For how many ordered triples <math>(x,y,z)</math> of nonnegative integers less than <math>20</math> are there exactly two distinct elements in the set <math>\{i^x, (1+i)^y, z\}</math>, where <math>i=\sqrt{-1}</math>?
 
For how many ordered triples <math>(x,y,z)</math> of nonnegative integers less than <math>20</math> are there exactly two distinct elements in the set <math>\{i^x, (1+i)^y, z\}</math>, where <math>i=\sqrt{-1}</math>?
  
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== Solution ==
 
== Solution ==
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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>.
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We have either <math>i^{x}=(1+i)^{y}\neq z</math>, <math>i^{x}=z\neq(1+i)^{y}</math>, or <math>(1+i)^{y}=z\neq i^x</math>.
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<math>i^{x}=(1+i)^{y}</math> only occurs when it is <math>1</math>. <math>(1+i)^{y}=1</math> has only one solution, namely, <math>y=0</math>. <math>i^{x}=1</math> has five solutions between 0 and 19, <math>x=0, x=4, x=8, x=12</math>, and <math>x=16</math>. <math>z\neq 1</math> has nineteen integer solutions between zero and nineteen. So for <math>i^{x}=(1+i)^{y}\neq z</math>, we have <math>5\cdot 1\cdot 19=95</math> ordered triples.
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For <math>i^{x}=z\neq(1+i)^{y}</math>, again this only occurs at <math>1</math>. <math>(1+i)^{y}\neq 1</math> has nineteen solutions, <math>i^{x}=1</math> has five solutions, and <math>z=1</math> has one solution, so again we have <math>5\cdot 1\cdot 19=95</math> ordered triples.
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For <math>(1+i)^{y}=z\neq i^x</math>, this occurs at <math>1</math> and <math>16</math>. <math>(1+i)^{y}=1</math> and <math>z=1</math> both have one solution while <math>i^{x}\neq 1</math> has fifteen solutions. <math>(1+i)^{y}=16</math> and <math>z=16</math> both have one solution, namely, <math>y=8</math> and <math>z=16</math>, while <math>i^{x}\neq 16</math> has twenty solutions (<math>i^x</math> only cycles as <math>1, i, -1, -i</math>). So we have <math>15\cdot 1\cdot 1+20\cdot 1\cdot 1=35</math> ordered triples.
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In total we have <math>{95+95+35=\boxed{\text{(D) }225}}</math> ordered triples
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== Small Clarification ==
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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 ==
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{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}
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{{AMC12 box|year=2010|num-b=14|num-a=16|ab=B}}
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{{MAA Notice}}

Latest revision as of 16:29, 26 May 2023

Problem

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 $i^{x}=(1+i)^{y}\neq z$, $i^{x}=z\neq(1+i)^{y}$, or $(1+i)^{y}=z\neq i^x$.

$i^{x}=(1+i)^{y}$ only occurs when it is $1$. $(1+i)^{y}=1$ has only one solution, namely, $y=0$. $i^{x}=1$ has five solutions between 0 and 19, $x=0, x=4, x=8, x=12$, and $x=16$. $z\neq 1$ has nineteen integer solutions between zero and nineteen. So for $i^{x}=(1+i)^{y}\neq z$, we have $5\cdot 1\cdot 19=95$ ordered triples.

For $i^{x}=z\neq(1+i)^{y}$, again this only occurs at $1$. $(1+i)^{y}\neq 1$ has nineteen solutions, $i^{x}=1$ has five solutions, and $z=1$ has one solution, so again we have $5\cdot 1\cdot 19=95$ ordered triples.

For $(1+i)^{y}=z\neq i^x$, this occurs at $1$ and $16$. $(1+i)^{y}=1$ and $z=1$ both have one solution while $i^{x}\neq 1$ has fifteen solutions. $(1+i)^{y}=16$ and $z=16$ both have one solution, namely, $y=8$ and $z=16$, while $i^{x}\neq 16$ has twenty solutions ($i^x$ only cycles as $1, i, -1, -i$). So we have $15\cdot 1\cdot 1+20\cdot 1\cdot 1=35$ ordered triples.

In total we have ${95+95+35=\boxed{\text{(D) }225}}$ 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 $x$, $y$ and $z$ respectively.

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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All AMC 12 Problems and Solutions

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