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}}</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>.
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== Video Solution by OmegaLearn (Using Probability) ==
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https://youtu.be/geEDrsV5Glw
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~ 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}}

Revision as of 14:55, 12 February 2021

Problem

Two fair dice, each with at least $6$ faces are rolled. On each face of each dice is printed a distinct integer from $1$ to the number of faces on that die, inclusive. The probability of rolling a sum if $7$ is $\frac34$ of the probability of rolling a sum of $10,$ and the probability of rolling a sum of $12$ is $\frac{1}{12}$. What is the least possible number of faces on the two dice combined?

$\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18\qquad \textbf{(D) }19 \qquad \textbf{(E) }20$

Solution

Suppose the dice have $a$ and $b$ faces, and WLOG $a\geq{b}$. Since each die has at least $6$ faces, there will always be $6$ ways to sum to $7$. As a result, there must be $\tfrac{4}{3}\cdot6=8$ ways to sum to $10$. There are at most nine distinct ways to get a sum of $10$, which are possible whenever $a,b\geq{9}$. To achieve exactly eight ways, $b$ must have $8$ faces, and $a\geq9$. Let $n$ be the number of ways to obtain a sum of $12$, then $\tfrac{n}{8a}=\tfrac{1}{12}\implies n=\tfrac{2}{3}a$. Since $b=8$, $n\leq8\implies a\leq{12}$. In addition to $3\mid{a}$, we only have to test $a=9,12$, of which both work. Taking the smaller one, our answer becomes $a+b=9+8=\boxed{\textbf{(B)17}}$.

Video Solution by OmegaLearn (Using Probability)

https://youtu.be/geEDrsV5Glw

~ pi_is_3.14

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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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