HMMT 十一月 2021 · GEN 赛 · 第 4 题

HMMT November 2021 — GEN Round — Problem 4

The sum of the digits of the time 19 minutes ago is two less than the sum of the digits of the time right now. Find the sum of the digits of the time in 19 minutes. (Here, we use a standard 12-hour clock of the form hh:mm.)

解析

The sum of the digits of the time 19 minutes ago is two less than the sum of the digits of the time right now. Find the sum of the digits of the time in 19 minutes. (Here, we use a standard 12-hour clock of the form hh:mm.) Proposed by: Holden Mui Answer: 11 Solution: Let’s say the time 19 minutes ago is h hours and m minutes, so the sum of the digits is equivalent to h + m mod 9. If m ≤ 40, then the time right now is h hours and m + 19 minutes, so the sum of digits is equivalent mod 9 to h + m + 19 ≡ h + m + 1 (mod 9), which is impossible. If m > 40 and h < 12, then the time right now is h + 1 hours and m − 41 minutes, so the sum of digits is equivalent to h + m − 40 ≡ h + m + 5 (mod 9), which is again impossible. Therefore, we know that h = 12 and m > 40. Now, the sum of the digits 19 minutes ago is 3 + s ( m ), where s ( n ) is the sum of the digits of n . On the other hand, the sum of the digits now is 1 + s ( m − 41), meaning that 4 + s ( m ) = s ( m − 41). The only m that satisfies this is m = 50, so the time right now is 1:09. In 19 minutes, the time will be 1:28, so the answer is 11.