2021 AMC 12B Problems/Problem 19
Contents
Problem
Two fair dice, each with at least faces are rolled. On each face of each dice is printed a distinct integer from
to the number of faces on that die, inclusive. The probability of rolling a sum of
is
of the probability of rolling a sum of
and the probability of rolling a sum of
is
. What is the least possible number of faces on the two dice combined?
Solution 1
Suppose the dice have and
faces, and WLOG
. Since each die has at least
faces, there will always be
ways to sum to
. As a result, there must be
ways to sum to
. There are at most nine distinct ways to get a sum of
, which are possible whenever
. To achieve exactly eight ways,
must have
faces, and
. Let
be the number of ways to obtain a sum of
, then
. Since
,
. In addition to
, we only have to test
, of which both work. Taking the smaller one, our answer becomes
.
Solution 2
Suppose the dice have and
faces, and WLOG
. Note that the amount of ways to make 12 is limited by how high the combined number of faces is since it cannot exceed 24 by the answer choices. This gives us the probability of making a 12 as
Cross-multiplying, we get
Simon's Favorite Factoring Trick now gives
This narrows the possibilities down to 3 ordered pairs of
, which are
,
, and
. We can obviously ignore the first pair and test the next two straightforwardly. The last pair yields the answer:
The answer is then
.
~Hyprox1413
Video Solution by OmegaLearn (Using Probability)
~ pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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