2021 AMC 12B Spring Problem 19

Question

Two fair dice, each with at least

6

faces are rolled. On each face of each die is printed a distinct integer from

1

to the number of faces on that die, inclusive. The probability of rolling a sum of

7

is

\dfrac34

of the probability of rolling a sum of

10,

and the probability of rolling a sum of

12

is

\dfrac{1}{12}.

What is the least possible number of faces on the two dice combined?

16

17

18

19

20

Answer: B

Concepts:dice (probability)optimization

Difficulty rating: 2120

Small hint:

With at least

6

faces each, a sum of

7

always has

6

ways, so a sum of

10

has

8

ways

Big hint:

Let the dice have

a\le b

faces; count outcomes for sums

10

and

12

and use

\text{ways}(12)=\tfrac{ab}{12}

Solution:

Let the dice have

a\le b

faces. Since both have at least

6

faces, a sum of

7

occurs in exactly

6

ways, so a sum of

10

occurs in

6\div\tfrac34=8

ways.

The number of ways to roll

10

is

\min(a,9)

-\max(1,10-b)+1=8.

A sum of

12

has probability

\tfrac{1}{12},

so it occurs in

\tfrac{ab}{12}

ways.

Having

8

outcomes for a sum of

10

requires

a\ge8

and

b\ge9,

so necessarily

a+b\ge17.

For

(a,b)=(8,9),

a sum of

10

has

8

outcomes, while a sum of

12

has the

6

outcomes with the first die showing

3,4,\ldots,8.

Since

\frac{6}{8\cdot9}=\frac{1}{12},

both conditions hold and the lower bound

17

is attained.

Thus, the correct answer is B.