Consider an n x n chessboard with all 4 corner squares removed. Prove that if the board can be covered with L-tetrominoes then n-2 is a multiple of 4. Is the converse true? (an L-tetromino is a plane figure shown below, constructed from four unit squares arranged in the form of L)

enter image description here


Hint: By counting the number of squares, $n$ is even.

Hint: By using the standard coloring of $i+j \pmod{4}$ for square $(i,j)$, show that $n \neq 4k$.

Hence $n= 4k+2$.

I'm not certain about the converse.

  • $\begingroup$ I'm not sure how to use your hints to solve the problem $\endgroup$ – Jebediah Nov 18 '13 at 7:14
  • $\begingroup$ @Jebediah How many squares are there in a $n\times n$ chessboard with 4 removed? How many squares are covered by an integer number of $L-$tetrominoes? Hence, show that $n$ is even. $\endgroup$ – Calvin Lin Nov 18 '13 at 15:03
  • $\begingroup$ There are $n^2 - 4$ squares after the removal. An L-tetromino covers a multiple of 4 squares per block. $\endgroup$ – Jebediah Nov 20 '13 at 15:04
  • $\begingroup$ @Jebediah Great. Do you see why that shows that $n^2-4$ must be a multiple of 4, and hence $n$ is even? $\endgroup$ – Calvin Lin Nov 20 '13 at 21:20
  • $\begingroup$ I don't see why $n^2-4$ must be a multiple of 4. Care to explain? $\endgroup$ – Jebediah Nov 21 '13 at 4:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.