2024 AMC 12B Problem 17

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}

If the roots are distinct integers

p, q, r,

then

pqr = -6,

a = -(p+q+r),

and

b = pq + qr + rp

List the ways to write

-6

as a product of three distinct integers, discard any giving

|a|\gt 10

or

|b| \gt 10,

and divide by

21 \cdot 20

ordered choices

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.

Problem 17 in other years