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

(Created page with "===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 al...")
 
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===Solution 1===
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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>.

Revision as of 19:33, 11 February 2021

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

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