A rook starts moving on an infinite chessboard, alternating horizontal and vertical moves. The length of the first move is one square, of the second – two squares, of the third – three squares and so on. a) Is it possible for the rook to arrive at its starting point after exactly $2013$ moves? b) Find all $n$ for which it possible for the rook to come back to its starting point after exactly $n$ moves.
Problem
Source: 2013 Romania NMO VIII p2
Tags: combinatorics, combinatorial geometry, Chessboard
29.09.2024 00:02
$a)$ Lets paint the board like it is a chess board: black and white. Then for every square that is black the square above, below, at the right and at the left are white. The same with the white squares. I denote $S(x)$ as the sum of the moves after the xth move. Then $S(x)=\frac{x(x+1)}{2}$. If the rook starts at a black square then at the nth square moved, where $n\equiv 0 \mod 2$ ,is black. After $2013$ moves the rook has moved $\frac{2013*2014}{2}=2027091$ squares (if it comes back to a square it counts like another square. Because $2027091 \equiv 1 \mod 2$ then after $2013$ moves the rock is not at a black square then it can not come back to the starting point.
06.11.2024 20:01
07.11.2024 00:35
a) it is impossible since it takes an even number of steps to get back to starting point