HMMT 二月 2026 · 团队赛 · 第 7 题
HMMT February 2026 — Team Round — Problem 7
题目详情
- [50] An infinite sequence a , a , a , . . . of integers is called r - close if every integer appears in the 1 2 3 sequence exactly once and | a − a | ≥ r | a − a | for all integers n ≥ 1 . Determine, with proof, n +2 n +1 n +1 n all nonnegative real numbers r for which an r -close sequence exists.
解析
- [50] An infinite sequence a , a , a , . . . of integers is called r - close if every integer appears in the 1 2 3 sequence exactly once and | a − a | ≥ r | a − a | for all integers n ≥ 1 . Determine, with proof, n +2 n +1 n +1 n all nonnegative real numbers r for which an r -close sequence exists. Proposed by: Derek Liu Answer: All real numbers r such that 0 ≤ r < 2 ©2026 HMMT − k Solution: First, we provide a construction when r < 2 . Let k be an integer such that r < 2 − 2 . We can enumerate the integers in the order { 0 , 1 , − 1 , 2 , − 2 , . . . } . To construct our sequence, we start with a = 0 and go through the enumeration in order. For each number m in the enumeration, if it is 1 already in the sequence, skip it. Otherwise, let the last term in the current sequence be a . n In the case where a > m , pick some positive integer d greater than twice the difference between any n two terms in the current sequence. Let a = a + d , a = a + 2 d , a = a + 4 d , and so on, n +1 n n +2 n +1 n +3 n +2 k up until a = a + 2 d . By definition of d , every previous term in the sequence is within d of n + k +1 n + k a , so none of a , a , . . . could have appeared previously. Note that the successive differences d , n n +1 n +2 2 d , 4 d , . . . are each double the last, with d being at least double the previous difference by definition; as r < 2 , this is valid. Furthermore, we can now let a = m , as n + k +2 k +1 a − m (2 − 1) d + ( a − m ) n + k +1 n − k = > 2 − 2 > r. k a − a 2 d n + k +1 n + k Now that we have added m to the sequence, we continue with the enumeration. In the case where a < m , pick some integer d in the similar manner but a = a − d , a = a − 2 d n n +1 n n +2 n +1 and so on. This similar construction allows us to add m to the sequence in this case too. Since every integer appears in the enumeration, every integer will appear in the sequence. By con- struction, every integer appears exactly once, as desired. It remains to show no such sequence exists for r ≥ 2 . Assume for sake of contradiction otherwise, and let m = a ± 1 , with sign chosen so that m ̸ = a . We know m must appear in the sequence somewhere, 2 1 say as a where n ≥ 3 . Note that | a − a | ≥ 2 | a − a | ≥ 2 , and by triangle inequality, n 3 2 2 1 | a − a | ≥ | a − a | − | a − a | − · · · − | a − a | n 2 n n − 1 n − 1 n − 2 3 2 ≥ | a − a | − · · · − | a − a | n − 1 n − 2 3 2 ≥ . . . ≥ | a − a | 3 2 = 2 , contradiction. Thus, such a sequence exists if and only if r < 2 . Remark. In the construction, we choose to enumerate the integers in this order: { 0 , 1 , − 1 , 2 , − 2 , . . . } . However, any enumerations that include all integers work.