PUMaC 2019 · 数论（A 组） · 第 4 题

PUMaC 2019 — Number Theory (Division A) — Problem 4

For a positive integer n , let f ( n ) = b log n c . Find the largest n < 2018 such that n | f ( n ). 2 i =1

解析

For a positive integer n , let f ( n ) = b log i c . Find the largest n < 2018 such that n | f ( n ). 2 i =1 Proposed by: Eric Neyman Answer: 1013 First note that r r r r ∑ ∑ ∑ ∑ r +1 k j r +1 i r +1 f (2 − 1) = k · 2 = 2 = (2 − 2 ) = ( r − 1)2 + 2 . k =0 i =1 j = i i =1 r +1 r +1 Thus, if we write n = 2 − 1 + m , where 0 ≤ m ≤ 2 , we have r +1 f ( n ) = ( r − 1)2 + 2 + m ( r + 1) . 1 Thus, the condition n | f ( n ) is equivalent (after subtracting ( r − 1) n from f ( n )) to r +1 2 − 1 + m | 2 + m ( r + 1) + r − 1 − m ( r − 1) = 2 m + r + 1 . Now, the right-hand side is more than zero times the left-hand side but more than twice the r +1 r +1 left-hand side, so n | f ( n ) if and only if 2 − 1 + m = 2 m + r + 1, i.e. m = 2 − r − 2, so r +2 n = 2 − r − 3. 10 The largest such value that is less than 2018 is 2 − 8 − 3 = 1013.