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?

$16$

$17$

$18$

$19$

$20$

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.

Trying $a=8,b=9$ : sum $10$ has $8-1+1=8$ ways, and sum $12$ has $6=\tfrac{72}{12}$ ways. Both conditions hold, giving $a+b=17.$

Checking all smaller totals $a+b=16$ fails, so $17$ is minimal.

Thus, the correct answer is B.

2021 AMC 12B Spring Problem 19

Solution:

Problem 19 in Other Years