Difference between revisions of "2024 AMC 12B Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
− | <math>-10 \ | + | Since <math>-10 \le a,b \le 10</math>, there are 21 integers to choose from, and <math>P(21,2) = 21 \times 20 = 420</math> equally likely ordered pairs <math>(a,b)</math>. |
− | Applying Vieta, | + | Applying Vieta's formulas, |
<math>x_1 \cdot x_2 \cdot x_3 = -6</math> | <math>x_1 \cdot x_2 \cdot x_3 = -6</math> | ||
− | <math> x_1 + x_2+ | + | <math> x_1 + x_2+ x_3 = -a </math> |
<math> x_1 \cdot x_2 + x_1 \cdot x_3 + x_3 \cdot x_2 = b</math> | <math> x_1 \cdot x_2 + x_1 \cdot x_3 + x_3 \cdot x_2 = b</math> | ||
Line 24: | Line 24: | ||
(2) <math> (x_1,x_2,x_3) = ( 1,2,-3) , b = -7, a=0</math> valid | (2) <math> (x_1,x_2,x_3) = ( 1,2,-3) , b = -7, a=0</math> valid | ||
− | (3) <math> (x_1,x_2,x_3) = (1,-2,3) , b = - | + | (3) <math> (x_1,x_2,x_3) = (1,-2,3) , b = -5, a=-2</math> valid |
− | (4) <math> (x_1,x_2,x_3) = (-1,2,3) , b = 1, a=4</math> valid | + | (4) <math> (x_1,x_2,x_3) = (-1,2,3) , b = 1, a=-4</math> valid |
(5) <math> (x_1,x_2,x_3) = (-1,-2,-3) , b = 11</math> invalid | (5) <math> (x_1,x_2,x_3) = (-1,-2,-3) , b = 11</math> invalid | ||
− | the total event space is <math>21 \cdot (21- 1)</math> (choice of select a | + | the total event space is <math>21 \cdot (21- 1)</math> (choice of select a times choice of selecting b given no-replacement) |
hence, our answer is <math>\frac{4}{21 \cdot 20} = \boxed{\textbf{(C) }\frac{1}{105}}</math> | hence, our answer is <math>\frac{4}{21 \cdot 20} = \boxed{\textbf{(C) }\frac{1}{105}}</math> | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | |||
+ | ==Solution 1.1 (desperation)== | ||
+ | As obtained in Solution 1, we get that there are <math>P(21,2) = 420</math> equally likely ordered pairs <math>(a,b)</math>, which means that the denominator will likely be a factor of 420, which leaves answers C and D, and if you are lucky enough, you can guess that the answer is <math>\boxed{\textbf{(C) }\frac{1}{105}}</math> | ||
+ | ~Soupboy0 | ||
+ | |||
+ | ==Video Solution 1 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=ptFW2866-Xw | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=B|num-b=16|num-a=18}} | {{AMC12 box|year=2024|ab=B|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:39, 21 November 2024
Contents
Problem 17
Integers and are randomly chosen without replacement from the set of integers with absolute value not exceeding . What is the probability that the polynomial has distinct integer roots?
.
Solution 1
Since , there are 21 integers to choose from, and equally likely ordered pairs .
Applying Vieta's formulas,
Cases:
(1) valid
(2) valid
(3) valid
(4) valid
(5) invalid
the total event space is (choice of select a times choice of selecting b given no-replacement)
hence, our answer is
Solution 1.1 (desperation)
As obtained in Solution 1, we get that there are equally likely ordered pairs , which means that the denominator will likely be a factor of 420, which leaves answers C and D, and if you are lucky enough, you can guess that the answer is ~Soupboy0
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=ptFW2866-Xw
See also
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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.