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Difference between revisions of "2021 AMC 12B Problems/Problem 19"

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<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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==Solution==
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==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>.
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==Solution 2==
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Suppose the dice have <math>a</math> and <math>b</math> faces, and WLOG <math>a\geq{b}</math>. 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 <cmath>\frac{a+b-11}{ab}=\frac{1}{12}</cmath>
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Cross-multiplying, we get
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<cmath>12a+12b-132=ab</cmath>
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Simon's Favorite Factoring Trick now gives
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<cmath>(a-12)(b-12)=12</cmath>
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This narrows the possibilities down to 3 ordered pairs of <math>(a,b)</math>, which are <math>(11,24)</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:
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<cmath>\frac{6}{72}=\frac{3}{4}\frac{9+8-9}{72}</cmath>
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The answer is then <math>a+b=8+9=\boxed{\textbf{(B)}\ 17}</math>.
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~Hyprox1413
  
 
== Video Solution by OmegaLearn (Using Probability) ==
 
== Video Solution by OmegaLearn (Using Probability) ==

Revision as of 14:28, 27 August 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 of $7$ 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 1

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 $a+b=9+8=\boxed{\textbf{(B)}\ 17}$.

Solution 2

Suppose the dice have $a$ and $b$ faces, and WLOG $a\geq{b}$. 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 \[\frac{a+b-11}{ab}=\frac{1}{12}\] Cross-multiplying, we get \[12a+12b-132=ab\] Simon's Favorite Factoring Trick now gives \[(a-12)(b-12)=12\] This narrows the possibilities down to 3 ordered pairs of $(a,b)$, which are $(11,24)$, $(6,10)$, and $(8,9)$. We can obviously ignore the first pair and test the next two straightforwardly. The last pair yields the answer: \[\frac{6}{72}=\frac{3}{4}\frac{9+8-9}{72}\] The answer is then $a+b=8+9=\boxed{\textbf{(B)}\ 17}$.

~Hyprox1413

Video Solution by OmegaLearn (Using Probability)

https://youtu.be/geEDrsV5Glw

~ 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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