Competition problem — Combinatorics

Library / /8 of 15

Combinatorics Difficulty 4.8 AIME Prove it United States

Problem:

Define a "word" to be a string of at most ten letters taken from the English alphabet. (The letters do not have to be distinct.) Prove that the number of "words" is divisible by $27$ .

Solution

Solution:

The total number of $9$ - and $10$ -letter words is divisible by $27$ , because they can be partitioned into groups of $27$ with each group containing a $9$ -letter word and the twenty-six $10$ -letter words formed by adding a letter at the end of it.

For the same reason, the number of $7$ - and $8$ -letter words is divisible by $27$ ; likewise for $5$ - and $6$ -, $3$ - and $4$ -, $1$ - and $2$ -letter words. Therefore the total number of words is divisible by $27$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.