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<= a, b <= 10 , a,b has 21 choices per Vieta, x1x2x3 = -6, x1 + x2+ x3 = -a , x1x2+ x2x3 + x3x1 = b

Case: (1) (x1,x2,x3) = (-1,-1,6) , b = 13 not valid

(2) (x1,x2,x3) = (-1,1,6) , b = -1, a=-6 valid

(3) (x1,x2,x3) = ( 1,2,-3) , b = -7, a=0 valid

(4) (x1,x2,x3) = (1,-2,3) , b = -7, a=2 valid

(5) (x1,x2,x3) = (-1,2,3) , b = 1, a=4 valid

(6) (x1,x2,x3) = (-1,-2,-3) , b = 11 invalid

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

~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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