HMMT 十一月 2024 · THM 赛 · 第 2 题

HMMT November 2024 — THM Round — Problem 2

Paul is in the desert and has a pile of gypsum crystals. No matter how he divides the pile into two nonempty piles, at least one of the resulting piles has a number of crystals that, when written in base 10 , has a sum of digits at least 7 . Given that Paul’s initial pile has at least two crystals, compute the smallest possible number of crystals in the initial pile.

解析

Paul is in the desert and has a pile of gypsum crystals. No matter how he divides the pile into two nonempty piles, at least one of the resulting piles has a number of crystals that, when written in base 10, has a sum of digits at least 7. Given that Paul’s initial pile has at least two crystals, compute the smallest possible number of crystals in the initial pile. Proposed by: Albert Wang, Carlos Rodriguez Answer: 49 Solution: Denote the digit sum of a positive integer m as s ( m ). i i i 1 2 s ( n ) Let the pile have n gypsum crystals, so that n can be written as 10 + 10 + · · · + 10 . First, s ( n ) cannot be 1 (i.e. n cannot be a power of 10), since otherwise we could split the gypsum pile into two equal parts each with digit sum 5. We also can’t have 1 < s ( n ) ≤ 12, as otherwise i i i 1 2 s ( n ) n = 10 + 10 + · · · + 10 can be split into two groups of at most six terms each (which have digit sum at most 6). Hence s ( n ) is at least 13, forcing n to be at least 49. Now we show that n = 49 indeed works. Note that for any splitting into two piles of size a , b with a + b = 49, we do not carry any digits when adding a and b . Hence we must have s ( a )+ s ( b ) = s ( n ) = 13. This forces at least one of s ( a ), s ( b ) to be at least 7, so n = 49 indeed works.