Difference between revisions of "2021 AMC 12B Problems/Problem 19"
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==Problem== | ==Problem== | ||
− | Two fair dice, each with at least <math>6</math> faces are rolled. On each face of each | + | Two fair dice, each with at least <math>6</math> faces are rolled. On each face of each die is printed a distinct integer from <math>1</math> to the number of faces on that die, inclusive. The probability of rolling a sum of <math>7</math> is <math>\frac34</math> of the probability of rolling a sum of <math>10,</math> and the probability of rolling a sum of <math>12</math> is <math>\frac{1}{12}</math>. What is the least possible number of faces on the two dice combined? |
<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 2== | ==Solution 2== | ||
− | Suppose the dice have <math>a</math> and <math>b</math> faces, and WLOG <math>a\geq{b}</math>. Note that | + | Suppose the dice have <math>a</math> and <math>b</math> faces, and WLOG <math>a\geq{b}</math>. Note that if <math>a+b=12</math> since they are both <math>6</math>, there is one way to make <math>12</math>, and incrementing <math>a</math> or <math>b</math> by one will add another way. This gives us the probability of making a 12 as |
+ | <cmath>\frac{a+b-11}{ab}=\frac{1}{12}</cmath> | ||
Cross-multiplying, we get | Cross-multiplying, we get | ||
<cmath>12a+12b-132=ab</cmath> | <cmath>12a+12b-132=ab</cmath> | ||
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<cmath>(a-12)(b-12)=12</cmath> | <cmath>(a-12)(b-12)=12</cmath> | ||
This narrows the possibilities down to 3 ordered pairs of <math>(a,b)</math>, which are <math>(13,24)</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: | This narrows the possibilities down to 3 ordered pairs of <math>(a,b)</math>, which are <math>(13,24)</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: | ||
− | <cmath>\frac{6}{72}=\frac{3}{4}\frac{9+8-9}{72}</cmath> | + | <cmath>\frac{6}{72}=\frac{3}{4}\left(\frac{9+8-9}{72}\right)</cmath> |
The answer is then <math>a+b=8+9=\boxed{\textbf{(B)}\ 17}</math>. | The answer is then <math>a+b=8+9=\boxed{\textbf{(B)}\ 17}</math>. | ||
~Hyprox1413 | ~Hyprox1413 | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/xGp5yQ5Bshs | ||
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+ | ~MathProblemSolvingSkills | ||
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== Video Solution by OmegaLearn (Using Probability) == | == Video Solution by OmegaLearn (Using Probability) == |
Latest revision as of 11:43, 27 May 2023
Contents
Problem
Two fair dice, each with at least faces are rolled. On each face of each die is printed a distinct integer from to the number of faces on that die, inclusive. The probability of rolling a sum of is of the probability of rolling a sum of and the probability of rolling a sum of is . What is the least possible number of faces on the two dice combined?
Solution 1
Suppose the dice have and faces, and WLOG . Since each die has at least faces, there will always be ways to sum to . As a result, there must be ways to sum to . There are at most nine distinct ways to get a sum of , which are possible whenever . To achieve exactly eight ways, must have faces, and . Let be the number of ways to obtain a sum of , then . Since , . In addition to , we only have to test , of which both work. Taking the smaller one, our answer becomes .
Solution 2
Suppose the dice have and faces, and WLOG . Note that if since they are both , there is one way to make , and incrementing or by one will add another way. This gives us the probability of making a 12 as Cross-multiplying, we get Simon's Favorite Factoring Trick now gives This narrows the possibilities down to 3 ordered pairs of , which are , , and . We can obviously ignore the first pair and test the next two straightforwardly. The last pair yields the answer: The answer is then .
~Hyprox1413
Video Solution
~MathProblemSolvingSkills
Video Solution by OmegaLearn (Using Probability)
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.