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Difference between revisions of "2010 AMC 12B Problems/Problem 15"
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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>.
 
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>.
  
−
For <math>i^{x}=(1+i)^{y}</math>, this only occurs at <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 zero and nineteen, <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\times 1\times 19=95</math> ordered triples.
+
For <math>i^{x}=(1+i)^{y}</math>, this only occurs at <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 zero and nineteen, <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.
  
−
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\times 1\times 19=95</math> ordered triples.
+
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.
  
−
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. So we have <math>15\times 1\times 1+20\times 1\times 1=35</math> ordered triples.
+
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.
  
−
In total we have <math>{95+95+35=225}</math> ordered triples <math>\Rightarrow \boxed{D}</math>
+
In total we have <math>{95+95+35=\boxed{\text{(D) }225}}</math> ordered triples
  
 
== 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 17:28, 9 October 2017

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

For $i^{x}=(1+i)^{y}$, this only occurs at $1$. $(1+i)^{y}=1$ has only one solution, namely, $y=0$. $i^{x}=1$ has five solutions between zero and nineteen, $x=0, x=4, x=8, x=12$, 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

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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