Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
Line 15: Line 15:
 
In total we have <math>{95+95+35=\boxed{\text{(D) }225}}</math> ordered triples
 
In total we have <math>{95+95+35=\boxed{\text{(D) }225}}</math> ordered triples
  
−
~ Small Clarification ~
+
==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 <math>x</math>, <math>y</math> and <math>z</math> respectively.
 
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.

Revision as of 16:24, 23 November 2020

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

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_15&oldid=138250"