Difference between revisions of "2021 AMC 12B Problems/Problem 19"
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==Solution== | ==Solution== | ||
| − | Suppose the dice have <math>a</math> and <math>b</math> faces, and WLOG <math>a\geq{b}</math>. Since each die has at least <math>6</math> faces, there will always be <math>6</math> ways to sum to <math>7</math>. As a result, there must be <math>\tfrac{4}{3}\cdot6=8</math> ways to sum to <math>10</math>. There are at most nine distinct ways to get a sum of <math>10</math>, which are possible whenever <math>a,b\geq{9}</math>. To achieve exactly eight ways, <math>b</math> must have <math>8</math> faces, and <math>a\geq9</math>. Let <math>n</math> be the number of ways to obtain a sum of <math>12</math>, then <math>\tfrac{n}{8a}=\tfrac{1}{12}\implies n=\tfrac{2}{3}a</math>. Since <math>b=8</math>, <math>n\leq8\implies a\leq{12}</math>. In addition to <math>3\mid{a}</math>, we only have to test <math>a=9,12</math>, of which both work. Taking the smaller one, our answer becomes <math>a+b=9+8=\boxed{\textbf{(B)17 | + | Suppose the dice have <math>a</math> and <math>b</math> faces, and WLOG <math>a\geq{b}</math>. Since each die has at least <math>6</math> faces, there will always be <math>6</math> ways to sum to <math>7</math>. As a result, there must be <math>\tfrac{4}{3}\cdot6=8</math> ways to sum to <math>10</math>. There are at most nine distinct ways to get a sum of <math>10</math>, which are possible whenever <math>a,b\geq{9}</math>. To achieve exactly eight ways, <math>b</math> must have <math>8</math> faces, and <math>a\geq9</math>. Let <math>n</math> be the number of ways to obtain a sum of <math>12</math>, then <math>\tfrac{n}{8a}=\tfrac{1}{12}\implies n=\tfrac{2}{3}a</math>. Since <math>b=8</math>, <math>n\leq8\implies a\leq{12}</math>. In addition to <math>3\mid{a}</math>, we only have to test <math>a=9,12</math>, of which both work. Taking the smaller one, our answer becomes <math>a+b=9+8=\boxed{\textbf{(B)}\ 17}</math>. |
== Video Solution by OmegaLearn (Using Probability) == | == Video Solution by OmegaLearn (Using Probability) == | ||
Revision as of 17:17, 3 March 2021
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 if 
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
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 
.
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. 
