2016 AIME I Problem 2

Attempt Problem 2 of the 2016 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AIME I solutions, or check the answer key.

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2.

Two dice appear to be standard dice with their faces numbered from 11 to 6,6, but each die is weighted so that the probability of rolling the number kk is directly proportional to k.k. The probability of rolling a 77 with this pair of dice is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Answer: 71
Concepts:dice (probability)basic probability
Difficulty rating: 2070
Solution:

Since 1+2++6=21,1 + 2 + \cdots + 6 = 21, each die rolls kk with probability k21.\frac{k}{21}. A total of 77 arises from the pairs (k,7k)(k, 7-k) for k=1,,6,k = 1, \ldots, 6, so its probability is 16+25+34+43+52+61212=56441=863. \begin{aligned} &\scriptsize \frac{1 \cdot 6 + 2 \cdot 5 + 3 \cdot 4 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1}{21^2} \\ &= \frac{56}{441} = \frac{8}{63}. \end{aligned}

Thus m+n=8+63=71.m + n = 8 + 63 = 71.

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