PUMaC 2015 · 代数（B 组） · 第 8 题
PUMaC 2015 — Algebra (Division B) — Problem 8
题目详情
- [ 8 ] We define the ridiculous numbers recursively as follows: (a) 1 is a ridiculous number. √ √ (b) If a is a ridiculous number, then a and 1 + a are also ridiculous numbers. A closed interval I is “boring” if • I contains no ridiculous numbers, and • There exists an interval [ b, c ] containing I for which b and c are both ridiculous numbers. The smallest non-negative l such that there does not exist a boring interval with length l can √ a + b c be represented in the form where a, b, c, d are integers, gcd( a, b, d ) = 1, and no integer d square greater than 1 divides c . What is a + b + c + d ? 1
解析
- [ 8 ] We define the ridiculous numbers recursively as follows: (a) 1 is a ridiculous number. √ √ (b) If a is a ridiculous number, then a and 1 + a are also ridiculous numbers. A closed interval I is “boring” if • I contains no ridiculous numbers, and • There exists an interval [ b, c ] containing I for which b and c are both ridiculous numbers. The smallest non-negative l such that there does not exist a boring interval with length l can √ a + b c be represented in the form where a, b, c, d are integers, gcd( a, b, d ) = 1, and no integer d square greater than 1 divides c . What is a + b + c + d ? Solution: √ √ The smallest ridiculous number is 1. This is true because if a ≥ 1, then a ≥ 1 and 1+ a ≥ 1. The supremum of the ridiculous numbers (the smallest number that is greater than all ridicu- √ 3+ 5 lous numbers) is . This is because the largest ridiculous number that is n recursive steps 2 √ √ √ away from 1 is 1 + 1 + 1 + ... 1, where there are n square root signs. As n approaches √ √ √ infinity, the largest ridiculous numbers approach M = 1 + 1 + 1 + 1 + ... . It can be seen √ √ 3+ 5 that M = by observing that M satisfies the identity M = 1 + M . 2 √ 1+ 5 There are no ridiculous numbers in the interval [ , 2): In the case that r is of the form 2 √ 1 + a for some ridiculous a , then r must be at least 2, because a must be at least 1. In the √ √ √ √ 3+ 5 1+ 5 case that r is of the form a , then r must be less than = . Since these are the 2 2 √ 1+ 5 only two cases, if I = [ , 2 − ] for some small > 0, then I is a boring interval, so l must 2 √ √ 1+ 5 3 − 5 be at least 2 − = . 2 2 √ 3 − 5 Now we show that there does not exist a boring interval of length . Assume for the sake 2 √ √ 1+ 5 3+ 5 of contradiction that such an interval exists. It must be contained in [1 , ] or [2 , ]. 2 2 √ √ √ The first case is ruled out because 2 and 2 are ridiculous numbers. The second case is √ √ √ √ 3 − 5 ruled out because 1 + 2 and 1 + 2 are ridiculous. Therefore l = and our answer is 2 3 + ( − 1) + 5 + 2 = 9 . Author: Ben Edelman 3