Question

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?

$\dfrac{1}{240}$

$\dfrac{1}{221}$

$\dfrac{1}{105}$

$\dfrac{1}{84}$

$\dfrac{1}{63}$

Solution:

The set has $21$ integers, so there are $21 \cdot 20 = 420$ ordered choices of $(a, b).$ If the polynomial has distinct integer roots $p, q, r,$ then $pqr = -6,$ $a = -(p + q + r),$ and $b = pq + qr + rp.$

The triples of distinct integers with product $-6$ are $\{1, 2, -3\},$ $\{1, -2, 3\},$ $\{-1, 2, 3\},$ $\{-1, -2, -3\},$ and $\{1, -1, 6\}.$ These give $(a, b) = (0, -7),$ $(-2, -5),$ $(-4, 1),$ $(6, 11),$ and $(-6, -1).$ The fourth has $b = 11 \gt 10,$ so it is invalid; the other four are valid and distinct.

The probability is $\dfrac{4}{420} = \dfrac{1}{105}.$

Thus, the correct answer is C.

2024 AMC 12B Problem 17

Solution:

Problem 17 in Other Years