2021 AMC 12B Problems/Problem 19

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 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

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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