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Difference between revisions of "2024 AMC 12B Problems/Problem 17"

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==Solution 1==
 
==Solution 1==
  
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<math>-10 \leq a, b \leq 10</math>, each of <math>a,b</math> has <math>21</math> choices
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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>.
  
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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>
  
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<math> x_1 + x_2+ x_2 = -a </math>
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<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>
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(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  
  
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(3)  <math> (x_1,x_2,x_3)  = (1,-2,3) , b = -7, a=2</math>  valid  
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(3)  <math> (x_1,x_2,x_3)  = (1,-2,3) , b = -5, a=-2</math>  valid  
  
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(4)  <math> (x_1,x_2,x_3)  = (-1,2,3) , b = 1, a=4</math>  valid  
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(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  
  
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the total event space is <math>21  \cdot (21- 1)</math> (choice of select a<math>\cdot</math>choice of selecting b given no-replacement)  
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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>
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~[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
 
==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}}

Revision as of 13:06, 20 November 2024

Problem 17

Integers $a$ and $b$ are randomly chosen without replacement from the set of integers with absolute value not exceeding $10$. What is the probability that the polynomial $x^3 + ax^2 + bx + 6$ has $3$ distinct integer roots?

$\textbf{(A) } \frac{1}{240} \qquad \textbf{(B) } \frac{1}{221} \qquad \textbf{(C) } \frac{1}{105} \qquad \textbf{(D) } \frac{1}{84} \qquad \textbf{(E) } \frac{1}{63}$.

Solution

Solution 1

Since $-10 \le a,b \le 10$, there are 21 integers to choose from, and $P(21,2) = 21 \times 20 = 420$ equally likely ordered pairs $(a,b)$.

Applying Vieta's formulas,

$x_1 \cdot x_2 \cdot x_3 = -6$

$x_1 + x_2+ x_3 = -a$

$x_1 \cdot x_2 + x_1 \cdot x_3 + x_3 \cdot x_2 = b$

Cases:

(1) $(x_1,x_2,x_3) = (-1,1,6) , b = -1, a=-6$ valid

(2) $(x_1,x_2,x_3) = ( 1,2,-3) , b = -7, a=0$ valid

(3) $(x_1,x_2,x_3) = (1,-2,3) , b = -5, a=-2$ valid

(4) $(x_1,x_2,x_3) = (-1,2,3) , b = 1, a=-4$ valid

(5) $(x_1,x_2,x_3) = (-1,-2,-3) , b = 11$ invalid

the total event space is $21 \cdot (21- 1)$ (choice of select a times choice of selecting b given no-replacement)

hence, our answer is $\frac{4}{21 \cdot 20} = \boxed{\textbf{(C) }\frac{1}{105}}$

~luckuso

Solution 1.1 (desperation)

As obtained in Solution 1, we get that there are $P(21,2) = 420$ equally likely ordered pairs $(a,b)$, 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 $\boxed{\textbf{(C) }\frac{1}{105}}$

~Soupboy0

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. AMC logo.png

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