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]]}} | ||
+ | |||
+ | ==Problem== | ||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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) | ||
+ | |||
==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.
Contents
Problem
When the roots of the polynomial
are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is positive?
Solution 1
is a product of
or 10 terms. When
, all terms are
, but
because there is an even number of terms. The sign keeps alternating
. There are 11 intervals, so there are
positives and 5 negatives.
~
Solution 2
Denote by the interval
for
and
the interval
.
Therefore, the number of intervals that is positive is
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.