HMMT 十一月 2010 · 冲刺赛 · 第 35 题
HMMT November 2010 — Guts Round — Problem 35
题目详情
- [ 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 descendents has had, according to the Mathematical Genealogy Project. Note that the Mathematical Genealogy Project has records dating back to the 1300s. 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 to 100 x .
解析
- [ 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 Round the number of descendents that the mathematician who has had the largest number of descendents has had, according to the Mathematical Genealogy Project. Note that the Mathematical Genealogy Project has records dating back to the 1300s. 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 to 100 x . Answer: 82310 First let’s estimate how many “generations” of mathematicians there have been since 1300. If we suppose that a mathematician gets his PhD around age 30 and becomes a PhD advisor around age 60, then we’ll get a generation length of approximately 30 years. However, not all mathematicians will train more than one PhD. Let’s say that only 40% of mathematicians train at least 2 PhDs. Then effectively we have only 40% of the generations, or in other words each effective 22 generation takes 75 years. Then we have branching generations. If we assume that all of these only 3 22 3 train 2 PhDs, then we get an answer of 2 ≈ 1625. But we can ensure that our chain has at least a single person who trained 100 PhDs (this is approximately the largest number of advisees for a single mathematician), 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.