HMMT 十一月 2021 · 冲刺赛 · 第 9 题

HMMT November 2021 — Guts Round — Problem 9

[7] Let n be an integer and m = ( n − 1001)( n − 2001)( n − 2002)( n − 3001)( n − 3002)( n − 3003) . Given that m is positive, find the minimum number of digits of m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2021, November 13, 2021 — GUTS ROUND Organization Team Team ID# 1

解析

[7] Let n be an integer and m = ( n − 1001)( n − 2001)( n − 2002)( n − 3001)( n − 3002)( n − 3003) . Given that m is positive, find the minimum number of digits of m . Proposed by: Daniel Zhu Answer: 11 Solution: One can show that if m > 0, then we must either have n > 3003 or n < 1001. If n < 1001, 5 each term other than n − 1001 has absolute value at least 1000, so m > 1000 , meaning that m has at least 16 digits. However, if n > 3003, it is clear that the minimal m is achieved at n = 3004, which makes m = 2002 · 1002 · 1001 · 3 · 2 · 1 = 12 · 1001 · 1001 · 1002 , 9 which is about 12 · 10 and thus has 11 digits. 1