HMMT 十一月 2010 · 冲刺赛 · 第 19 题

HMMT November 2010 — Guts Round — Problem 19

[ 11 ] How many 8-digit numbers begin with 1, end with 3, and have the property that each successive digit is either one more or two more than the previous digit, considering 0 to be one more than 9?

解析

[ 11 ] How many 8-digit numbers begin with 1, end with 3, and have the property that each successive digit is either one more or two more than the previous digit, considering 0 to be one more than 9? Guts Round Answer: 21 Given an 8-digit number a that satifies the conditions in the problem, let a denote i the difference between its ( i + 1)th and i th digit. Since i ∈ { 1 , 2 } for all 1 ≤ i ≤ 7, we have 7 ≤ a + a + · · · + a ≤ 14. The difference between the last digit and the first digit of m is 3 − 1 ≡ 2 1 2 7 (mod 10), which means a + · · · + a = 12. Thus, exactly five of the a s equal to 2 and the remaining 1 7 i ( ) 7 two equal to 1. The number of permutations of five 2s and two 1s is = 21. 2