2024 AMC 12B Problems/Problem 17

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

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2024 AMC 12B Problems/Problem 17

Problem 17

Solution 1

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