HMMT 十一月 2020 · THM 赛 · 第 9 题
HMMT November 2020 — THM Round — Problem 9
题目详情
- While waiting for their next class on Killian Court, Alesha and Belinda both write the same sequence S on a piece of paper, where S is a 2020-term strictly increasing geometric sequence with an integer common ratio r . Every second, Alesha erases the two smallest terms on her paper and replaces them with their geometric mean, while Belinda erases the two largest terms in her paper and replaces them with their geometric mean. They continue this process until Alesha is left with a single value A and Belinda is left with a single value B . Let r be the minimal value of r 0 A such that is an integer. If d is the number of positive factors of r , what is the closest integer to log d ? 0 2 B
解析
- While waiting for their next class on Killian Court, Alesha and Belinda both write the same sequence S on a piece of paper, where S is a 2020-term strictly increasing geometric sequence with an integer common ratio r . Every second, Alesha erases the two smallest terms on her paper and replaces them with their geometric mean, while Belinda erases the two largest terms in her paper and replaces them with their geometric mean. They continue this process until Alesha is left with a single value A and A Belinda is left with a single value B . Let r be the minimal value of r such that is an integer. If d 0 B is the number of positive factors of r , what is the closest integer to log d ? 0 2 Proposed by: Hahn Lheem Answer: 2018 Solution: Because we only care about when the ratio of A to B is an integer, the value of the first 2 2019 term in S does not matter. Let the initial term in S be 1. Then, we can write S as 1 , r, r , . . . , r . a b Because all terms are in terms of r , we can write A = r and B = r . We will now solve for a and b . m + n m n 2 Observe that the geometric mean of two terms r and r is simply r , or r raised to the arithmetic mean of m and n . Thus, to solve for a , we can simply consider the sequence 0 , 1 , 2 , . . . , 2019, which comes from the exponents of the terms in S , and repeatedly replace the smallest two terms with their arithmetic mean. Likewise, to solve for b , we can consider the same sequence 0 , 1 , 2 , . . . , 2019 and repeatedly replace the largest two terms with their arithmetic mean. We begin by computing a . If we start with the sequence 0 , 1 , . . . , 2019 and repeatedly take the arith- metic mean of the two smallest terms, the final value will be 0+1 +2 2 +3 2019 2 ∑
- · · · + 2019 k 2 a = = . 2020 − k 2 2 k =1 Then, we can compute 2019 ∑ k 2 a = 2019 − k 2 k =1 2019 2019 ∑ ∑ k k = ⇒ a = 2 a − a = − 2019 − k 2020 − k 2 2 k =1 k =1 2019 2018 ∑ ∑ k k + 1 = − 2019 − k 2019 − k 2 2 k =1 k =0 2019 ∑ 1 = 2019 − j 2 j =1 ( ) 1 1 = 2019 − 1 − = 2018 + . 2019 2019 2 2 1 Likewise, or by symmetry, we can find b = 1 − . 2019 2 ( ) ( ) a A r 1 1 a − b Since we want = = r to be a positive integer, and a − b = 2018 + − 1 − = b 2019 2019 B r 2 2 1 2018 th 2017 + , r must be a perfect (2 ) power. Because r > 1, the minimal possible value is 2018 2 2018 2 2018 r = 2 . Thus, d = 2 + 1, and so log d is clearly closest to 2018. 2