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

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


<math>\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}</math>.
<math>\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}</math>.


[[2024 AMC 12B Problems/Problem 17|Solution]]
[[2024 AMC 12B Problems/Problem 17|Solution]]

Revision as of 22:56, 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 invalid

(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

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

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

~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

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