HMMT 十一月 2011 · 冲刺赛 · 第 24 题
HMMT November 2011 — Guts Round — Problem 24
题目详情
- [ 12 ] Three not necessarily distinct positive integers between 1 and 99, inclusive, are written in a row on a blackboard. Then, the numbers, without including any leading zeros, are concatenated to form a new integer N . For example, if the integers written, in order, are 25, 6, and 12, then N = 25612 (and not N = 250612). Determine the number of possible values of N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TH 4 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND 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 be the answer to problem 27.
解析
- [ 12 ] Three not necessarily distinct positive integers between 1 and 99, inclusive, are written in a row on a blackboard. Then, the numbers, without including any leading zeros, are concatenated to form a new integer N . For example, if the integers written, in order, are 25, 6, and 12, then N = 25612 (and not 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 for 3 each. So we have a total of 90 = 729000 possibilities. • Case 2: 5 digits. Then, exactly one of the three numbers is between 1 and 9, inclusive. We consider cases on the presence of 0s in N . 5 – No 0s. Then, we have 9 choices for each digit, for a total of 9 = 59049 choices. – One 0. Then, the 0 can be the second, third, fourth, or fifth digit, and 9 choices for each of 4 the other 4 digits. Then, we have a total of 4 × 9 = 26244 choices. – Two 0s. Then, there must be at least one digit between them and they cannot be in the first 3 digit, giving us 3 choices for the positioning of the 0s. Then, we have a total of 3 ∗ 9 = 2187 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 0s. 4 – No 0s. Then, we have 9 = 6561 choices. 3 – One 0. Then, the 0 can go in the second, third, or fourth digit, so we have 3 × 9 = 2187 choices. So we have a total of 6561 + 2187 = 8748 choices in this case. 3 • Case 4: 3 digits. Then, we cannot have any 0s, so we have a total of 9 = 729 choices. Hence, we have a total of 729000 + 87480 + 8748 + 729 = 825957 choices for N . Guts Round