2021 AMC 12B Problems/Problem 19

Problem

Two fair dice, each with at least $6$ faces are rolled. On each face of each die 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 if $a+b=12$ since they are both $6$ , there is one way to make $12$ , and incrementing $a$ or $b$ by one will add another way. 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 $(13,24)$ , $(6,10)$ , and $(8,9)$ . We can obviously ignore the first pair and test the next two straightforwardly. The last pair yields the answer: $\frac{6}{72}=\frac{3}{4}\left(\frac{9+8-9}{72}\right)$ The answer is then $a+b=8+9=\boxed{\textbf{(B)}\ 17}$ .

~Hyprox1413

Solution 3 (Logic)

Notice that $\begin{align*} 7&=1+6 \\ &\ \ \ \ \ \ \ \ \ \vdots \\ &=6+1 \end{align*}$ are the only cases that could possibly form 7. In another words, regardless of the number of faces for each dice, 6 is the number of cases that could form a rolling sum of 7.

Because the value of the total combination for both are the same, we could infer that the number of cases of having a rolling sum of 10 is $\frac{4}{3}\cdot6=8$ . With the number 8, we could also deduce that one dice has 8 sides and the other has at least 9 sides. Thence trial and error could be utilized.

Since 9 is the next smallest number, the case could be tested. $\begin{align*} 12&=3+9 \\ &=4+8 \\ &\ \ \ \ \ \ \ \ \ \vdots \\ &=8+4 \end{align*}$ $\frac{6}{8\cdot9}=\frac{1}{12}$ is also true. Therefore, $8+9=\boxed{\textbf{(B) }17}$

~MaPhyCom

Solution 4

Let the two dice have at least six faces each, and let their face counts sum to $N$ . We seek the minimum possible $N$ such that $p(\text{sum}=12)=\tfrac{1}{12}$ and $p(\text{sum}=7)=\tfrac{3}{4}p(\text{sum}=10)$ .

First try $N=16$ . Then the possible face-count pairs $(m,n)$ with $m \leq n$ are $(6,10)$ , $(7,9)$ , $(8,8)$ . For $p(\text{sum}=12)=\tfrac{1}{12}$ to be possible, the total outcomes $mn$ must be a multiple of 12. But $7\cdot 9=63$ and $8\cdot 8=64$ are not multiples of 12, so $(7,9)$ and $(8,8)$ are impossible.

For $(6,10)$ , there are 60 outcomes. The combinations that sum to 12 are (2,10), (3,9), (4,8), (5,7), (6,6), five in total, giving $p(12)=\tfrac{5}{60}=\tfrac{1}{12}$ . The combinations that sum to 10 are (1,9), (2,8), (3,7), (4,6), (5,5), (6,4), six in total, giving $p(10)=\tfrac{6}{60}=\tfrac{1}{10}$ . The combinations that sum to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), also six in total, giving $p(7)=\tfrac{6}{60}=\tfrac{1}{10}$ . So $p(7)=p(10)$ , which does not satisfy the condition. Hence $N=16$ is impossible.

Next try $N=17$ . The possible pairs are $(6,11)$ , $(7,10)$ , $(8,9)$ . Again, $mn$ must be a multiple of 12, but $6\cdot 11=66$ and $7\cdot 10=70$ are not; discard these.

For $(8,9)$ , there are 72 outcomes. The combinations that sum to 12 are (3,9), (4,8), (5,7), (6,6), (7,5), (8,4), six in total, so $p(12)=\tfrac{6}{72}=\tfrac{1}{12}$ . The combinations that sum to 10 are (1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), eight in total, so $p(10)=\tfrac{8}{72}=\tfrac{1}{9}$ . The combinations that sum to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), six in total, so $p(7)=\tfrac{6}{72}=\tfrac{1}{12}$ .

Thus $p(12)=\tfrac{1}{12}$ and $p(7)=\tfrac{1}{12}=\tfrac{3}{4}\cdot\tfrac{1}{9}=\tfrac{3}{4}p(10)$ .

Therefore $N=17$ works, and since $N=16$ is impossible, the minimum is $\boxed{(\text{B})\17}$ .

~STIDE

Video Solution

https://youtu.be/xGp5yQ5Bshs

~MathProblemSolvingSkills

Video Solution by OmegaLearn (Using Probability)

https://youtu.be/geEDrsV5Glw

~ pi_is_3.14

2021 AMC 12B (Problems • Answer Key • Resources)
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