HMMT 二月 2003 · 冲刺赛 · 第 40 题
HMMT February 2003 — Guts Round — Problem 40
题目详情
英文原题
- [18] All the sequences consisting of five letters from the set { T, U, R, N, I, P } (withrepetitions allowed) are arranged in alphabetical order in a dictionary. Two sequencesare called “anagrams” of each other if one can be obtained by rearranging the letters ofthe other. How many pairs of anagrams are there that have exactly 100 other sequencesbetween them in the dictionary?
解析
英文解析
- All the sequences consisting of five letters from the set { T, U, R, N, I, P } (with repetitions allowed) are arranged in alphabetical order in a dictionary. Two sequences arecalled “anagrams” of each other if one can be obtained by rearranging the letters of the other. How many pairs of anagrams are there that have exactly 100 other sequences 11
between them in the dictionary?
Solution: 0
Convert each letter to a digit in base 6: I 7 → 0 , N 7 → 1 , P 7 → 2 , R 7 → 3 , T 7 → 4 , U 7 → 5.
Then the dictionary simply consists of all base-6 integers from 00000 to 55555 in
6 6
numerical order. If one number can be obtained from another by a rearrangement ofdigits, then the numbers are congruent modulo 5 (this holds because a number abcde
4 3 26 = 6 · a + 6 · b + 6 · c + 6 · d + e is congruent modulo 5 to a + b + c + d + e ), but ifthere are 100 other numbers between them, then their difference is 101, which is notdivisible by 5. So there are no such pairs.