MathDB
Unique word !

Source: India TST 2001 Day 1 Problem 2

January 31, 2015
inequalitiescombinatorics proposedcombinatorics

Problem Statement

Two symbols AA and BB obey the rule ABBB=BABBB = B. Given a word x1x2x3n+1x_1x_2\ldots x_{3n+1} consisting of nn letters AA and 2n+12n+1 letters BB, show that there is a unique cyclic permutation of this word which reduces to BB.