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Difference between revisions of "2023 AMC 10B Problems/Problem 12"

(→‎Solution)
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~Technodoggo
 
~Technodoggo
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 +
==Solution==
 +
 +
Denote by <math>I_k</math> the interval <math>\left( k - 1 , k \right)</math> for <math>k \in \left\{ 2, 3, \cdots , 10 \right\}</math> and <math>I_1</math> the interval <math>\left( - \infty, 1 \right)</math>.
 +
 +
Therefore, the number of intervals that <math>P(x)</math> is positive is
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 17:09, 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

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.

~Technodoggo

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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