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> | ||
− | ==Solution== | + | ==Solution 1== |
− | 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>. | + | 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>. |
+ | |||
+ | ==Solution 2== | ||
+ | Suppose the dice have <math>a</math> and <math>b</math> faces, and WLOG <math>a\geq{b}</math>. Note that if <math>a+b=12</math> since they are both <math>6</math>, there is one way to make <math>12</math>, and incrementing <math>a</math> or <math>b</math> by one will add another way. This gives us the probability of making a 12 as | ||
+ | <cmath>\frac{a+b-11}{ab}=\frac{1}{12}</cmath> | ||
+ | Cross-multiplying, we get | ||
+ | <cmath>12a+12b-132=ab</cmath> | ||
+ | Simon's Favorite Factoring Trick now gives | ||
+ | <cmath>(a-12)(b-12)=12</cmath> | ||
+ | This narrows the possibilities down to 3 ordered pairs of <math>(a,b)</math>, which are <math>(13,24)</math>, <math>(6,10)</math>, and <math>(8,9)</math>. We can obviously ignore the first pair and test the next two straightforwardly. The last pair yields the answer: | ||
+ | <cmath>\frac{6}{72}=\frac{3}{4}\left(\frac{9+8-9}{72}\right)</cmath> | ||
+ | The answer is then <math>a+b=8+9=\boxed{\textbf{(B)}\ 17}</math>. | ||
+ | |||
+ | ~Hyprox1413 | ||
+ | |||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/xGp5yQ5Bshs | ||
+ | |||
+ | ~MathProblemSolvingSkills | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | == Video Solution by OmegaLearn (Using Probability) == | ||
+ | https://youtu.be/geEDrsV5Glw | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2021|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:43, 27 May 2023
Contents
[hide]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 |
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