HMMT 十一月 2017 · THM 赛 · 第 4 题

HMMT November 2017 — THM Round — Problem 4

M ary has a sequence m , m , m , . . . , such that for each b ≥ 2, m is the least positive integer m for 2 3 4 b which none of the base- b logarithms log ( m ) , log ( m + 1) , . . . , log ( m + 2017) are integers. Find the b b b largest number in her sequence.

解析

M ary has a sequence m , m , m , . . . , such that for each b ≥ 2, m is the least positive integer m for 2 3 4 b which none of the base- b logarithms log ( m ) , log ( m + 1) , . . . , log ( m + 2017) are integers. Find the b b b largest number in her sequence. Proposed by: Michael Tang Answer: 2188 It is not difficult to see that for all of the logarithms to be non-integers, they must lie strictly between n +1 n n n and n + 1 for some integer n . Therefore, we require b − b > 2018, and so m = b + 1 where n b n n − 1 is the smallest integer that satisfies the inequality. In particular, this means that b − b ≤ 2018. 11 12 11 7 Note that m = 2 + 1 = 2049 (since 2 − 2 = 2048 > 2018) and m = 3 + 1 = 2188 (since 2 3 8 7 3 − 3 = 4374 > 2018). we now show that 2188 is the maximum possible value for m . b If n = 0, then m = 1 + 1 = 2. b If n = 1, then b − 1 ≤ 2018 and thus m = b + 1 ≤ 2020. b 2 2 If n = 2, then b − b ≤ 2018, which gives b ≤ 45, and thus m = b + 1 ≤ 2018 + b + 1 ≤ 2065. b 3 2 3 3 If n = 3, then b − b ≤ 2018, which gives b ≤ 12, and thus m = b + 1 ≤ 12 + 1 = 1729. b 4 3 4 4 If n = 4, then b − b ≤ 2018, which gives b ≤ 6, and thus m = b + 1 ≤ 6 + 1 = 1297. b 5 4 It then remains to check the value of m and m . Indeed, m = 4 + 1 = 1025 and m = 5 + 1 = 626, 4 5 4 5 so no values of m exceeds 2188. b