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2024 AMC 12B Problems/Problem 17: Difference between revisions

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(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</math> (21- 1)<math> (choice of select a</math>\cdot<math>choice of selecting b given no-replacement)  
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)  


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

Revision as of 18:28, 16 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,

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

$x_1 + x_2+ x_2 = -a$

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

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) $(x_1,x_2,x_3) = (1,-2,3) , b = -7, a=2$ 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$\cdot$choice of selecting b given no-replacement)

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

~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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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