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| − | When the roots of the polynomial
| + | #redirect[[2023 AMC 12B Problems/Problem 6]] |
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| − | <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?
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| − | ==Solution 1==
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| − | <math>P(x)</math> is a product of <math>(x-r_n)</math> or 10 terms. When <math>x < 1</math>, all terms are <math>< 0</math>, but <math>P(x) > 0</math> because there is an even number of terms. The sign keeps alternating <math>+,-,+,-,....,+</math>. There are 11 intervals, so there are <math>\boxed{\textbf{ 6}}</math> positives and 5 negatives. <math>\boxed{\textbf{(C) 6}}</math>
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| − | ~<math>\textbf{Techno}\textcolor{red}{doggo}</math>
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| − | ==Solution==
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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\{
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| − | \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)
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