PUMaC 2019 · 组合（B 组） · 第 4 题

PUMaC 2019 — Combinatorics (Division B) — Problem 4

Keith has 10 coins labeled 1 through 10, where the i th coin has weight 2 . The coins are all 1 fair, so the probability of flipping heads on any of the coins is . After flipping all of the 2 coins, Keith takes all of the coins which land heads and measures their total weight, W . If the probability that 137 ≤ W ≤ 1061 is m/n for coprime positive integers m, n , determine m + n .

解析

Keith has 10 coins labeled 1 through 10, where the i th coin has weight 2 . The coins are all 1 fair, so the probability of flipping heads on any of the coins is . After flipping all of the 2 1 coins, Keith takes all of the coins which land heads and measures their total weight, W . If the probability that 137 ≤ W ≤ 1061 is m/n for coprime positive integers m, n , determine m + n . Proposed by Alan Yan. Answer: 743 . Solution: We note that these weights form binary numbers, except the ”1” is omitted. Thus the numbers that are generated are exactly the even numbers between 2 and 2046, inclusive. Thus the number of possibilities is the number of even numbers between 138 and 1060, inclusive, 10 which is exactly 462. There are 2 = 1024 possible weights, which gives us a probability of 462 / 1024 = 231 / 512 for an answer of 743.