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Revision as of 12:56, 31 May 2011

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, $i^{x}=1$ has five solutions between zero and nineteen, and $z\neq 1$ 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 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 $\Rightarrow \boxed{D}$

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
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