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HMMT 十一月 2011 · 冲刺赛 · 第 24 题

HMMT November 2011 — Guts Round — Problem 24

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [ 12 ] Three not necessarily distinct positive integers between 1 and 99, inclusive, are written in a rowon a blackboard. Then, the numbers, without including any leading zeros, are concatenated to form anew integer N . For example, if the integers written, in order, are 25, 6, and 12, then N = 25612 (andnot N = 250612). Determine the number of possible values of N .
    4 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUNDTH
    Round 9
    The answers to the following three problems are mutually dependent, although your answer to each will be graded independently. Let A be the answer to problem 25, B the answer to problem 26, and C bethe answer to problem 27.

英文原题

[12]
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4 TH ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011— GUTS ROUND
School Team Team ID#

解析

英文解析

  1. [ 12 ] Three not necessarily distinct positive integers between 1 and 99, inclusive, are written in a rowon a blackboard. Then, the numbers, without including any leading zeros, are concatenated to form anew integer N . For example, if the integers written, in order, are 25, 6, and 12, then N = 25612 (andnot N = 250612). Determine the number of possible values of N .
    Answer: 825957 We will divide this into cases based on the number of digits of N .
    • Case 1: 6 digits. Then each of the three numbers must have two digits, so we have 90 choices foreach. So we have a total of 90 = 729000 possibilities.3
    • Case 2: 5 digits. Then, exactly one of the three numbers is between 1 and 9, inclusive. Weconsider cases on the presence of 0 s in N .
    – No 0 s. Then, we have 9 choices for each digit, for a total of 9 = 59049 choices.5
    – One 0. Then, the 0 can be the second, third, fourth, or fifth digit, and 9 choices for each ofthe other 4 digits. Then, we have a total of 4 × 9 = 26244 choices.4
    – Two 0 s. Then, there must be at least one digit between them and they cannot be in the firstdigit, giving us 3 choices for the positioning of the 0 s. Then, we have a total of 3 ∗ 9 = 21873
    choices.
    So we have a total of 59049 + 26244 + 2187 = 87480 choices in this case.
    • Case 3: 4 digits. Again, we casework on the presence of 0 s.
    – No 0 s. Then, we have 9 = 6561 choices.4
    – One 0. Then, the 0 can go in the second, third, or fourth digit, so we have 3 × 9 = 21873
    choices.
    So we have a total of 6561 + 2187 = 8748 choices in this case.
    • Case 4: 3 digits. Then, we cannot have any 0 s, so we have a total of 9 = 729 choices.3
    Hence, we have a total of 729000 + 87480 + 8748 + 729 = 825957 choices for N .
    Guts Round