2023 AMC 10B Problems/Problem 12

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

The interval of the alternating signs. $P(x)$ is a product of $(x-r_n)$ or 10 terms. When $x < 1$ , all terms are $< 0$ , but $P(x) > 0$ because they are even number of terms. The sign keep alternates $+,-,+,-,....,+$ . There are 11 intervals, so there are 6 positives and 5 negatives.

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

Solution

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)

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2023 AMC 10B Problems/Problem 12

Solution 1

Solution