2023 AMC 12B Problems/Problem 6

The following problem is from both the 2023 AMC 10B #12 and 2023 AMC 12B #6, so both problems redirect to this page.

Problem

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. $\boxed{\textbf{(C) 6}}$

~ $\textbf{Techno}\textcolor{red}{doggo}$

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)

Solution 3

We can use the turning point behavior at the roots of a polynomial graph to find out the amount of intervals that are positive.

First, we evaluate any value on the interval $(-\infty, 1)$ . Since the degree of $P(x)$ is $1+2+...+9$ = $\frac{9\times10}{2}$ = $45$ , and every term in P(x) is negative, multiplying 45 negatives gives a negative value. So $(-\infty, 0)$ is a negative interval.

We know that the roots of P(x) are at $1,2,...,10$ . When the degree of the term of each root is odd, the graph of P(x) will pass through the graph and change signs, and vice versa. So at $x=1$ , the graph will change signs; at $x=2$ , the graph will not, and so on.

This tells us that the interval $(1,2)$ is positive, $(2,3)$ is also positive, $(3,4)$ is negative, $(4,5)$ is also negative, and so on, with the pattern being $+,+,-,-,+,+,-,-,...$ .

The positive intervals are therefore $(1,2)$ , $(2,3)$ , $(5,6)$ , $(6,7)$ , $(9,10)$ , and $(10,\infty)$ , for a total of $\boxed{\textbf{(C) 6}}$ .

~nm1728

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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 10 Problems and Solutions

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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.

Page

Toolbox

Search

2023 AMC 12B Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also