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 "2021 Fall AMC 12B Problems/Problem 14"

m (→‎Solution)
m (→‎See Also)
Line 15: Line 15:
  
 
==See Also==
 
==See Also==
−
{{AMC12 box|year=2021 Fall|ab=B|num-a=13|num-b=11}}
+
{{AMC12 box|year=2021 Fall|ab=B|num-a=15|num-b=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:14, 16 January 2022

Problem

Suppose that $P(z), Q(z)$, and $R(z)$ are polynomials with real coefficients, having degrees $2$, $3$, and $6$, respectively, and constant terms $1$, $2$, and $3$, respectively. Let $N$ be the number of distinct complex numbers $z$ that satisfy the equation $P(z) \cdot Q(z)=R(z)$. What is the minimum possible value of $N$?

$\textbf{(A)}\: 0\qquad\textbf{(B)} \: 1\qquad\textbf{(C)} \: 2\qquad\textbf{(D)} \: 3\qquad\textbf{(E)} \: 5$

Solution

The answer cannot be $0$, as every nonconstant polynomial has at least $1$ distinct complex root (fundamental theorem of algebra); the polynomial $P(z)\cdot Q(z)$ has degree $2 + 3 = 5$, so the polynomial $R(z) - P(z)\cdot Q(z)$ has degree $6$ and is thus nonconstant. It now suffices to illustrate an example for which $N = 1$. Take $P(z)=z^2+1,Q(z)=z^3+2,$ and $R(z)=(z+1)^6 + \left(z^2+1\right)\left(z^3+2\right).$

$R(z)$ has degree 6 and constant term $3$, so it satisfies the conditions. We need to find the solutions to \[\left(z^2+1\right)\left(z^3+2\right)=(z+1)^6 + \left(z^2+1\right)\left(z^3+2\right),\] or \[(z+1)^6=0.\] Clearly, there is one distinct complex root, $-1$, so our answer is $\boxed{\textbf{(B)} \: 1}.$


~kingofpineapplz and kgator

See Also

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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=2021_Fall_AMC_12B_Problems/Problem_14&oldid=170044"