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HMMT 十一月 2010 · 冲刺赛 · 第 35 题

HMMT November 2010 — Guts Round — Problem 35

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 25 ] A mathematician M is called a descendent of mathematician M if there is a sequence of math-
    ′
    ematicians M = M , M , . . . , M = M such that M was M ’s doctoral advisor for all i . Estimate
    1 2 k i i +1
    the number of descendents that the mathematician who has had the largest number of descendentshas had, according to the Mathematical Genealogy Project. Note that the Mathematical Genealogy
    Project has records dating back to the 1300 s. If the correct answer is X and you write down A , your
    ( )
    | X − A |
    team will receive max 25 − b c , 0 points, where b x c is the largest integer less than or equal tox .100
解析

英文解析

  1. [ 25 ] A mathematician M is called a descendent of mathematician M if there is a sequence of math-
    ′
    ematicians M = M , M , . . . , M = M such that M was M ’s doctoral advisor for all i . Estimate
    1 2 k i i +1
    Guts Roundthe number of descendents that the mathematician who has had the largest number of descendentshas had, according to the Mathematical Genealogy Project. Note that the Mathematical Genealogy
    Project has records dating back to the 1300 s. If the correct answer is X and you write down A , your
    ( )
    | X − A |
    team will receive max 25 − b c , 0 points, where b x c is the largest integer less than or equal tox .100
    Answer: 82310 First let’s estimate how many “generations” of mathematicians there have beensince 1300. If we suppose that a mathematician gets his Ph d around age 30 and becomes a Ph dadvisor around age 60, then we’ll get a generation length of approximately 30 years. However, not allmathematicians will train more than one Ph d. Let’s say that only 40% of mathematicians train atleast 2 Ph ds. Then effectively we have only 40% of the generations, or in other words each effectivegeneration takes 75 years. Then we have branching generations. If we assume that all of these only 22
    3223
    train 2 Ph ds, then we get an answer of 2 ≈ 1625. But we can ensure that our chain has at least asingle person who trained 100 Ph ds (this is approximately the largest number of advisees for a singlemathematician), allowing us to change one factor of 2 into a factor of 100. That gives us an answer of
    1625 · 50 = 81250, which is very close to the actual value of 82310.