# 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 '15 at 8:58
• If you want a solvable challenge, use these pieces to build a 12x16 rectangle (there will be one left over). – Mike Earnest May 28 '15 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' – James Webster May 29 '15 at 7:23
• en.wikipedia.org/wiki/… – RobEarl May 29 '15 at 14:21
• You've got an odd number of "T"s, so no, not possible.... Oh, I see @Tryth already go it. – RBarryYoung May 29 '15 at 15:06

It is impossible.

Proof:
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. – Mike Earnest May 27 '15 at 7:08
• Very reminiscent of the classic "dominoes on a chess board with the corners removed" proof. I like it. – undergroundmonorail May 27 '15 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. – dberm22 May 27 '15 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. – Ross Millikan May 27 '15 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 '15 at 15:44

## protected by leoll2Jun 2 '15 at 16:11

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