Q:

Dividing square of 2013x2013

Is the square of 2013×2013 can be divided into rectangles 1×3 in such
  a way that the number of rectangles arranged vertically differ by 1
  from the number rectangles arranged horizontally? Prove your answer

I totally have no idea how to solve that. I need accurate solutions or good hints. 

A:

Since $2013$ is divisble by $3$, show by complete induction that the number of vertical bars that extend upward starting with bottom base at row $k$ must be divisible by $3$. Then you will get that the total number of vertical bars is divisible by $3$, and by symmetry, the number of horizontal bars that extend rightward starting with base column $k$ must be divisible by $3$, so the total number of horizontal bars must also be divisible by $3$, so they cannot differ by $1$.

