HMMT 十一月 2020 · 冲刺赛 · 第 31 题

HMMT November 2020 — Guts Round — Problem 31

[17] For some positive real α , the set S of positive real numbers x with { x } > αx consists of the union of a several intervals, with total length 20 . 2. The value of α can be expressed as , where a, b are relatively b prime positive integers. Compute 100 a + b . (Here, { x } = x − b x c is the fractional part of x .)

解析

[17] For some positive real α , the set S of positive real numbers x with { x } > αx consists of the a union of several intervals, with total length 20 . 2. The value of α can be expressed as , where a, b are b relatively prime positive integers. Compute 100 a + b . (Here, { x } = x − b x c is the fractional part of x .) Proposed by: Daniel Zhu Answer: 4633 α Solution: If we note that x = { x } + b x c , then we can rewrite our given inequality as { x } > b x c . 1 − α α However, since { x } < 1, we know that we must have b x c < { x } < 1, so each interval is of the form 1 − α 1 − ( n +1) α α ( n + n, n + 1) for some integer n , which has length . If we let k be the smallest integer 1 − α 1 − α 1 − ( k +1) α such that < 0, then the total length of all our intervals is the sum 1 − α k − 1 k ( k +1) ∑ k − α 1 − ( n + 1) α 2 = . 1 − α 1 − α n =0 If we set this to 20 . 2, we can solve for α to get k − 20 . 2 α = . k ( k +1) − 20 . 2 2 Since we defined k to be the smallest integer such that 1 − ( k + 1) α < 0, we know that k is the largest integer such that kα < 1. If we plug in our value for α , we get that this is equivalent to 2 k − 20 . 2 k < 1 = ⇒ k < 40 . 4 . k ( k +1) − 20 . 2 2 Thus, we have k = 40, and plugging this in for our formula for α gives us 40 − 20 . 2 33 α = = . 40 · 41 1333 − 20 . 2 2