2021 AMC 12B Problems/Problem 19
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[hide]Problem
Two fair dice, each with at least faces are rolled. On each face of each dice 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 |
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All AMC 12 Problems and Solutions |
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