Difference between revisions of "2021 AMC 12B Problems/Problem 19"
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==Problem== | ==Problem== | ||
− | Two fair dice, each with at least <math>6</math> faces are rolled. On each face of each | + | Two fair dice, each with at least <math>6</math> faces are rolled. On each face of each die is printed a distinct integer from <math>1</math> to the number of faces on that die, inclusive. The probability of rolling a sum of <math>7</math> is <math>\frac34</math> of the probability of rolling a sum of <math>10,</math> and the probability of rolling a sum of <math>12</math> is <math>\frac{1}{12}</math>. What is the least possible number of faces on the two dice combined? |
<math>\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18\qquad \textbf{(D) }19 \qquad \textbf{(E) }20</math> | <math>\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18\qquad \textbf{(D) }19 \qquad \textbf{(E) }20</math> | ||
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~Hyprox1413 | ~Hyprox1413 | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/xGp5yQ5Bshs | ||
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+ | ~MathProblemSolvingSkills | ||
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== Video Solution by OmegaLearn (Using Probability) == | == Video Solution by OmegaLearn (Using Probability) == |
Latest revision as of 12:43, 27 May 2023
Contents
Problem
Two fair dice, each with at least faces are rolled. On each face of each die 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 if
since they are both
, there is one way to make
, and incrementing
or
by one will add another way. 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
~MathProblemSolvingSkills
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.