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Difference between revisions of "2023 AMC 12B Problems/Problem 6"

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{{duplicate|[[2023 AMC 10B Problems/Problem 12|2023 AMC 10B #12]] and [[2023 AMC 12B Problems/Problem 6|2023 AMC 12B #6]]}}
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==Problem==
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When the roots of the polynomial
 +
 +
<math>P(x)  = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}</math>
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 +
are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is <math>P(x)</math> positive?
 +
 
==Solution 1==
 
==Solution 1==
  
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~<math>\textbf{Techno}\textcolor{red}{doggo}</math>
 
~<math>\textbf{Techno}\textcolor{red}{doggo}</math>
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 +
==Solution 2==
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 +
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>.
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Therefore, the number of intervals that <math>P(x)</math> is positive is
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<cmath>
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\begin{align*}
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1 + \sum_{i=1}^{10} \Bbb I \left\{
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\sum_{j=i}^{10} j \mbox{ is even}
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\right\}
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& = 1 + \sum_{i=1}^{10} \Bbb I \left\{
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\frac{\left( i + 10 \right) \left( 11 - i \right)}{2} \mbox{ is even}
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\right\} \\
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& = 1 + \sum_{i=1}^{10} \Bbb I \left\{
 +
\frac{- i^2 + i + 110}{2} \mbox{ is even}
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\right\} \\
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& = 1 + \sum_{i=1}^{10} \Bbb I \left\{
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\frac{i^2 - i}{2} \mbox{ is odd}
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\right\} \\
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& = \boxed{\textbf{(C) 6}} .
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\end{align*}
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</cmath>
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 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
  
 
==See Also==
 
==See Also==
 +
{{AMC10 box|year=2023|ab=B|num-b=11|num-a=13}}
 
{{AMC12 box|year=2023|ab=B|num-b=5|num-a=7}}
 
{{AMC12 box|year=2023|ab=B|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:32, 15 November 2023

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)


See Also

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

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