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

(→‎Solution 1)
(→‎Solution 1)
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==Solution 1==
 
==Solution 1==
  
−
-10<= a, b <= 10 , a,b has 21 choices
+
-10 <math>\leq</math> a, b <math>\leq</math> 10 , each of a,b has 21 choices
−
per Vieta, x1x2x3 = -6, x1 + x2+ x3 = -a , x1x2+ x2x3 + x3x1 = b
+
 
 +
Applying Vieta,  
 +
 
 +
<math>x_1 \cdot x_2  \cdot x_3 = -6</math>
 +
 
 +
<math> x_1 + x_2+ x_2 = -a </math>
 +
 
 +
<math> x_1 \cdot x_2 + x_1 \cdot x_3  + x_3 \cdot x_2  = b</math>
  
 
Case:
 
Case:
−
(1)  (x1,x2,x3) = (-1,-1,6) , b = 13 not valid
 
  
−
(2)  (x1,x2,x3) = (-1,1,6) , b = -1, a=-6  valid  
+
(1)  <math> (x_1,x_2,x_3) </math> = (-1,-1,6) , b = 13 not valid  
  
−
(3)  (x1,x2,x3) = ( 1,2,-3) , b = -7, a=0 valid  
+
(2)  <math> (x_1,x_2,x_3) </math> = (-1,1,6) , b = -1, a=-6 valid  
  
−
(4)  (x1,x2,x3) = (1,-2,3) , b = -7, a=2 valid  
+
(3)  <math> (x_1,x_2,x_3) </math> = ( 1,2,-3) , b = -7, a=0 valid  
  
−
(5)  (x1,x2,x3) = (-1,2,3) , b = 1, a=4 valid  
+
(4)  <math> (x_1,x_2,x_3) </math> = (1,-2,3) , b = -7, a=2 valid  
  
−
(6)  (x1,x2,x3) = (-1,-2,-3) , b = 11 invalid
+
(5)  <math> (x_1,x_2,x_3) </math> = (-1,2,3) , b = 1, a=4  valid
  
−
probability = <math>\frac{4}{21*20}</math> = <math>\frac{1}{105}</math> <math>\boxed{\textbf{(C) }\frac{1}{105}}</math>
+
(6)  <math> (x_1,x_2,x_3) </math> = (-1,-2,-3) , b = 11 invalid
 +
 
 +
probability = <math>\frac{4}{21*20}</math> =  <math>\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]
 +
 
==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 22:18, 14 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

-10 $\leq$ a, b $\leq$ 10 , each of a,b has 21 choices

Applying Vieta,

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

$x_1 + x_2+ x_2 = -a$

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

Case:

(1) $(x_1,x_2,x_3)$ = (-1,-1,6) , b = 13 not valid

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

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

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

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

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

probability = $\frac{4}{21*20}$ = $\boxed{\textbf{(C) }\frac{1}{105}}$

~luckuso

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

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