2011 AMC 12A Problem 14

Question

Suppose

a

and

b

are single-digit positive integers chosen independently and at random. What is the probability that the point

(a, b)

lies above the parabola

y = ax^2 - bx?

\dfrac{11}{81}

\dfrac{13}{81}

\dfrac{5}{27}

\dfrac{17}{81}

\dfrac{19}{81}

The point is above the parabola when

b \gt a \cdot a^2 - b \cdot a,

i.e.

b(a + 1) \gt a^3

Count valid

b

in

\{1, \ldots, 9\}

for each

a;

only

a = 1, 2, 3

allow any

Substituting

x = a,

y = b,

the point is above the parabola when

b \gt a^3 - ab,

i.e.

b(a + 1) \gt a^3.

For

a = 1:

b \gt \tfrac12,

all

9

values work. For

a = 2:

b \gt \tfrac83,

so

b \ge 3,

giving

7.

For

a = 3:

b \gt \tfrac{27}{4} = 6.75,

so

b \ge 7,

giving

3.

For

a \ge 4,

no

b \le 9

works.

The count is

9 + 7 + 3 = 19

out of

81,

so the probability is

\dfrac{19}{81}.

Thus, the correct answer is E.

Problem 14 in other years