# Is this Tetris puzzle solvable?

As a birthday present last year, I received some fridge magnets. They didn't come as a puzzle, so I don't know if they have a solution, but I made a puzzle out of them anyway.

The magnets are tetrominoes. There are 7 of each shape. Is it possible to arrange them into a 7x28 rectangle so that they are all used and all inside the rectangle?

The closest I have managed is this:

• As an aside, just because it's a puzzle doesn't mean it has a solution; the 15 puzzle, for example, had a popularized unsolvable configuration.
– user1502
May 28, 2015 at 8:58
• If you want a solvable challenge, use these pieces to build a 12x16 rectangle (there will be one left over). May 28, 2015 at 14:22
• Thank you to the upvoters! This question has just become my biggest SE achievement as 'the highest voted question on the site' May 29, 2015 at 7:23
• en.wikipedia.org/wiki/… May 29, 2015 at 14:21
• You've got an odd number of "T"s, so no, not possible.... Oh, I see @Tryth already go it. May 29, 2015 at 15:06

## 1 Answer

It is impossible.

Let the $$7\times 28$$ area be painted with black and white squares in a checkerboard pattern. Every piece will cover $$2$$ black and $$2$$ white squares, except the T-piece, which covers $$3$$ of one color and $$1$$ of another. Since there are $$7$$ T-pieces, a tiling that uses every piece cannot cover the same number of black and white squares. Since the board contains the same number of black and white squares, it is impossible.

• Nice! Also shows that you can't cover a 4x49 or 14x14 board. May 27, 2015 at 7:08
• Very reminiscent of the classic "dominoes on a chess board with the corners removed" proof. I like it. May 27, 2015 at 12:56
• @undergroundmonorail is referring to the Mutilated Chessboard Problem (en.wikipedia.org/wiki/Mutilated_chessboard_problem). It's not all corners which are removed, just the 2 opposing (catty-corner) corners (which are the same color). The solution hinges on there being the same number of white boxes as there are black, as does this proof. May 27, 2015 at 14:46
• @Zibbobz: Sure. Once you can make one rectangle with $n=4$ you can just line up as many of those rectangles as you want. May 27, 2015 at 16:40
• @kangacam How could the T piece ever do anything other than 3 and 1? It's a single square (the center of the T piece) with three squares all adjacent to that one single square. Thus the single square in its center must always be a different color than the three other squares by definition. The zigzag pieces on the other hand must cover 2 and 2: their center two pieces are always opposite colors, and each has one other piece adjacent which also must be an opposite color.
– Joe
May 28, 2015 at 15:44