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HMMT 二月 1998 · ADV 赛 · 第 9 题

HMMT February 1998 — ADV Round — Problem 9

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Bob’s Rice ID number has six digits, each a number from 1 to 9, and any digit can be used any number of times. The ID number satifies the following property: the first two digits is a number divisible by 2, the first three digits is a number divisible by 3, etc. so that the ID number itself is divisible by 6. One ID number that satifies this condition is 123252. How many different possibilities are there for Bob’s ID number?
解析
  1. Bob’s Rice ID number has six digits, each a number from 1 to 9, and any digit can be used any number of times. The ID number satisfies the following property: the first two digits is a number divisible by 2, the first three digits is a number divisible by 3, etc. so that the ID number itself is divisible by 6. One ID number that satisfies this condition is 123252. How many different possibilities are there for Bob’s ID number? Answer: 324 . We will count the number of possibilities for each digit in Bob’s ID number, then multiply them to find the total number of possibilities for Bob’s ID number. There are 3 possibilities for the first digit given any last 5 digits, because the entire number must be divisible by 3, so the sum of the digits must be divisible by 3. Because the first two digits are a number divisible by 2, the second digit must be 2, 4, 6, or 8, which is 4 possibilities. Because the first five digits are a number divisible by 5, the fifth digit must be a 5. Now, if the fourth digit is a 2, then the last digit has two choices, 2 , 8, and the third digit has 5 choices, 1 , 3 , 5 , 7 , 9. If the fourth digit is a 4, then the last digit must be a 6, and the third digit has 4 choices, 2 , 4 , 6 , 8. If the fourth digit is a 6, then the last digit must be a 4, and the third digit has 5 choices, 1 , 3 , 5 , 7 , 9. If the fourth digit is an 8, then the last digit has two choices, 2 , 8, and the third digit has 4 choices, 2 , 4 , 6 , 8. So there are a total of 3 · 4 (2 · 5 + 4 + 5 + 2 · 4) = 3 · 4 · 27 = 324 possibilities for Bob’s ID number.