HMMT 十一月 2019 · 团队赛 · 第 5 题

HMMT November 2019 — Team Round — Problem 5

[40] Compute the sum of all positive real numbers x ≤ 5 satisfying 2 d x e + d x e · b x c x = . d x e + b x c

解析

[40] Compute the sum of all positive real numbers x ≤ 5 satisfying 2 d x e + d x e · b x c x = . d x e + b x c Proposed by: Milan Haiman Answer: 85 k Note that all integer x work. If x is not an integer then suppose n < x < n + 1 . Then x = n + , where 2 n +1 n is an integer and 1 ≤ k ≤ 2 n is also an integer, since the denominator of the fraction on the right hand side is 2 n + 1 . We now show that all x of this form work. Note that ( ) ( ) 2 2 2 nk k k k 2 2 2 x = n + + = n + k − + . 2 n + 1 2 n + 1 2 n + 1 2 n + 1 ( ) 2 k k k 1 2 2 2 For between 0 and 1 , − + is between − and 0, so we have n + k − 1 < x ≤ n + k, 2 n +1 2 n +1 2 n +1 4 2 2 and d x e = n + k. Then, 2 2 d x e + d x e · b x c n + k + n · ( n + 1) k = = n + = x, d x e + b x c 2 n + 1 2 n + 1 so all x of this form work. Now, note that the 2 n solutions in the interval ( n, n + 1) , together with the solution n + 1 , form an n +1 arithmetic progression with 2 n + 1 terms and average value n + . Thus, the sum of the solutions in 2 n +1 2 2 2 the interval ( n, n + 1] is 2 n + 2 n + 1 = n + ( n + 1) . Summing this for n from 0 to 4 , we get that the answer is 2 2 2 2 2 2 0 + 2(1 + 2 + 3 + 4 ) + 5 = 85 .