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 "2023 AMC 10B Problems/Problem 12"

(→‎Solution 2)
(→‎Solution 1)
Line 9: Line 9:
 
<math>P(x)</math> is a product of <math>(x-r_n)</math> or 10 terms. When <math>x < 1</math>, all terms are <math>< 0</math>, but <math>P(x) > 0</math> because there is an even number of terms. The sign keeps alternating <math>+,-,+,-,....,+</math>.  There are 11 intervals, so there are <math>\boxed{\textbf{6}}</math> positives and 5 negatives. <math>\boxed{\textbf{(C) 6}}</math>
 
<math>P(x)</math> is a product of <math>(x-r_n)</math> or 10 terms. When <math>x < 1</math>, all terms are <math>< 0</math>, but <math>P(x) > 0</math> because there is an even number of terms. The sign keeps alternating <math>+,-,+,-,....,+</math>.  There are 11 intervals, so there are <math>\boxed{\textbf{6}}</math> positives and 5 negatives. <math>\boxed{\textbf{(C) 6}}</math>
  
−
~<math>\textbf{Techno}\textcolor{red}{doggo}</math>
+
~Technodoggo
  
 
==Solution 2==
 
==Solution 2==

Revision as of 18:59, 15 November 2023

When the roots of the polynomial

$P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}$

are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is $P(x)$ positive?

Solution 1

$P(x)$ is a product of $(x-r_n)$ or 10 terms. When $x < 1$, all terms are $< 0$, but $P(x) > 0$ because there is an even number of terms. The sign keeps alternating $+,-,+,-,....,+$. There are 11 intervals, so there are $\boxed{\textbf{6}}$ positives and 5 negatives. $\boxed{\textbf{(C) 6}}$

~Technodoggo

Solution 2

Denote by $I_k$ the interval $\left( k - 1 , k \right)$ for $k \in \left\{ 2, 3, \cdots , 10 \right\}$ and $I_1$ the interval $\left( - \infty, 1 \right)$.

Therefore, the number of intervals that $P(x)$ is positive is \begin{align*} 1 + \sum_{i=1}^{10} \Bbb I \left\{ \sum_{j=i}^{10} j \mbox{ is even} \right\} & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{\left( i + 10 \right) \left( 11 - i \right)}{2} \mbox{ is even} \right\} \\ & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{- i^2 + i + 110}{2} \mbox{ is even} \right\} \\ & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{i^2 - i}{2} \mbox{ is odd} \right\} \\ & = \boxed{\textbf{(C) 6}} . \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10B_Problems/Problem_12&oldid=203480"