2010 AMC 12B Problems/Problem 15

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 $5\times 1\times 19=95$ 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 $5\times 1\times 19=95$ 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. So we have $15\times 1\times 1+20\times 1\times 1=35$ ordered triples.

In total we have ${95+95+35=225}$ ordered triples $\Rightarrow \boxed{D}$

2010 AMC 12B (Problems • Answer Key • Resources)
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2010 AMC 12B Problems/Problem 15

Problem 15

Solution

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