# Is it possible to cover an $11 \times 12$ rectangle with $19$ rectangles of $1 \times 6$ or $1 \times 7$?

Is it possible to cover an $$11 \times 12$$ rectangle with $$19$$ rectangles of $$1 \times 6$$ or $$1 \times 7$$?

Attempt:

There should be $$132$$ unit squares to be covered. Since there are $$19$$ rectangles to be used, let $$x$$ be the number of $$1 \times 6$$ rectangles and $$19-x$$ be the number of $$1 \times 7$$ rectangles. The solution of $$(19-x)7 + (x)6 = 132$$ is $$133-7x + 6x = 132 \implies x = 1$$

So there should be only $$1$$ rectangle of $$1 \times 6$$, and $$18$$ rectangles of $$1 \times 7$$.

Now color the $$132$$ unit squares black and white like a chessboard, the top left is black..then its right is white...then its right is black again.. and so on.

Odd rows should be $$[black]-[white]-[black]- ... -[black]-[white]-[black]$$ Even rows is should be $$[white]-[black]-[white]- ... -[black]-[white]-[black]$$

For a $$1 \times 6$$ rectangle, it will definitely cover $$3$$ blacks and $$3$$ whites.

For a $$1 \times 7$$ rectangle, it will either cover $$4$$ blacks and $$3$$ whites, or cover $$3$$ blacks and $$4$$ whites. Let the number of $$4$$ blacks-$$3$$ whites coverings be $$y$$, and $$18-y$$ for the other one.

Notice that in total we must have $$66$$ blacks and $$66$$ whites. So in total there will be $$|black \: squares| = 3 + 4y + 3(18-y) = 57 + y \implies y = 9$$ $$|white \: squares| = 3 + 3y + 4(18-y) = 75 - y \implies y = 9$$

So there should be $$9$$ rectangles that cover $$4$$ blacks-$$3$$ whites, and $$9$$ rectangles that cover $$3$$ blacks- $$4$$ whites.

$$\matrix{\color{red}1,0,0,0,0,0,0,\color{red}1,0,0,0,0\\0,\color{red}1,0,0,0,0,0,0,\color{red}1,0,0,0\\0,0,\color{red}1,0,0,0,0,0,0,\color{red}1,0,0\\0,0,0,\color{red}1,0,0,0,0,0,0,\color{red}1,0\\0,0,0,0,\color{red}1,0,0,0,0,0,0,\color{red}1\\0,0,0,0,0,\color{red}1,0,0,0,0,0,0\\0,0,0,0,0,0,\color{red}1,0,0,0,0,0\\\color{red}1,0,0,0,0,0,0,\color{red}1,0,0,0,0\\0,\color{red}1,0,0,0,0,0,0,\color{red}1,0,0,0\\0,0,\color{red}1,0,0,0,0,0,0,\color{red}1,0,0\\0,0,0,\color{red}1,0,0,0,0,0,0,\color{red}1,0}$$

We have $$20$$ red cells and $$19$$ rectangles. Each rectangle can cover at most $$1$$ red cell. So..

• Not very deep, yet elegant :-) Commented Oct 22, 2019 at 9:22
• This is really nice approach and I also upvoted but shouldn't one also consider what the remaining pieces are? For instance if we have two $1 \times 3$ rectangles instead of a $1 \times 6$ rectangle, we can cover the board although we have $19$ rectangles and $20$ red cells. Commented Oct 22, 2019 at 10:16
• @ArsenBerk wouldn't that be 20 rectangles then?
– Sten
Commented Oct 22, 2019 at 14:16
• Ah, yes you are right. I had to say a $2 \times 3$ rectangle instead of two $1 \times 3$ rectangles. But in that case, since it could cover two red cells, that wouldn't be a problem so don't mind my question at all :D Commented Oct 22, 2019 at 15:03

Same as @Aqua's answer but I made a pretty image:

No two rectangles can cover more than one red square, so you need at least $$20$$ rectangles to cover all red squares.