2006 AIME II Problem 5


When rolling a certain unfair six-sided die with faces numbered 1,2,3,4,51, 2, 3, 4, 5, and 66, the probability of obtaining face FF is greater than 16\frac{1}{6}, the probability of obtaining the face opposite is less than 16\frac{1}{6}, the probability of obtaining any one of the other four faces is 16\frac{1}{6}, and the sum of the numbers on opposite faces is 77. When two such dice are rolled, the probability of obtaining a sum of 77 is 47288\frac{47}{288}. Given that the probability of obtaining face FF is mn\frac{m}{n}, where mm and nn are relatively prime positive integers, find m+nm+n.


Leading zeroes must be inputted, so if your answer is 34, then input 034. Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.

Show/Hide Hints

Show/Hide Problem Tags

Problem Tags: Algebra Counting and probability

Want to contribute problems and receive full credit? Click here to add your problem!
Please report any issues to us in our Discord server
Go to previous contest problem (SHIFT + Left Arrow) Go to next contest problem (SHIFT + Right Arrow)
Category: AIME II
Points: 3
Back to practice