HMMT 十一月 2020 · 团队赛 · 第 1 题

HMMT November 2020 — Team Round — Problem 1

[20] For how many positive integers n ≤ 1000 does the equation in real numbers b x c x = n have a positive solution for x ? (For a real number x , b x c denotes the largest integer that is not greater than x .)

解析

[20] For how many positive integers n ≤ 1000 does the equation in real numbers b x c x = n have a positive solution for x ? (For a real number x , b x c denotes the largest integer that is not greater than x .) Proposed by: John Michael Wu Answer: 412 b x c b 1 c Solution: If b x c = 0, then x = 1 = 1 , so we can safely ignore this case, as it does not introduce new solutions. b x c k If b x c = k for some k > 0, x 7 → x = x is a continuous and increasing function on the interval b x c k k 4 [ k, k + 1). Therefore, the x can take on any value in [ k , ( k + 1) ) when b x c = k . Because 5 < 5 k k 1000 < 5 , it suffices to count the number of integers in the interval [ k , ( k + 1) ) for 1 ≤ k ≤ 4. Thus, the number of valid n is simply 4 ∑ k k ( k + 1) − k = (2 − 1) + (9 − 4) + (64 − 27) + (625 − 256) = 412 . k =1