# 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?

$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\qquad\textbf{(E)}~5$

## Solution 1

The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about $(x-1)^1(x-3)^3(x-5)^5(x-7)^7(x-9)^9$. We need 0, 2, or 4 of the expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 through 6 and 6 through 7 intervals make two of the expressions negative. The 1 through 2 and 2 through 3 intervals make four of the expressions negative. There are $\boxed{\textbf{(C) 6}}$ intervals.

~Aopsthedude

## Solution 2

The roots of the factorized polynomial are intervals from numbers 1 to 10. We take each interval as being defined as the number behind it. To make the function positive, we need to have an even number of negative expressions. Real numbers raised to even powers are always positive, so we only focus on $x-1$, $x-3$, $x-5$, $x-7$, and $x-9$. The intervals 1 and 2 leave 4 negative expressions, so they are counted. The same goes for intervals 5, 6, 9, and 10. Intervals 3 and 4 leave 3 negative expressions and intervals 7 and 8 leave 1 negative expression. The solution is the number of intervals which is $\boxed{\textbf{(C) 6}}$.

~darrenn.cp ~DarkPheonix

## 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 $1+2+...+9+10$ = $\frac{10\times11}{2}$ = $55$, and every term in $P(x)$ is negative, multiplying $55$ 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 ~ESAOPS (minor edits)

## Solution 4

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)

## Video Solution

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