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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:

enter image description here

Rand al'Thor
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James Webster
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1 Answers1

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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.

Hazel へいぜる
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Tryth
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    Nice! Also shows that you can't cover a 4x49 or 14x14 board. – Mike Earnest May 27 '15 at 07:08
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    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
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    @undergroundmonorail is referring to the Mutilated Chessboard Problem (https://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
  • Is there a multiplier of n where n equals the number of tetrominoes of each type where the puzzle is solvable? – Zibbobz May 27 '15 at 15:00
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    @Zibbobz $n=4$ is doable. – Mike Earnest May 27 '15 at 15:33
  • @MikeEarnest Would the same hold true for n = 4x? Where x is the set of whole positive numbers? – Zibbobz May 27 '15 at 15:34
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    @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
  • Am I correct in reasoning that since the total number of squares on the board is even, all potential rectangular boards for $n=7$ would have an equal number of black and white squares, and so there is no rectangular board that works for $n=7$? – jpmc26 May 28 '15 at 00:23
  • In response to @Tryth's answer, the T piece will cover two and two if it is placed at the 'top' of a square. To be precise, 2 in the square it is in, 1 in the alternate colour above, and 1 in the alternate colour to the left or right - making 2 of each colour. The zig-zag / lightning bolt piece can only do 3 and 1, but there are 14 of those pieces. So is it still unsolvable? –  May 28 '15 at 07:29
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    @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
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    This is the same approach used for the Tetris lamp proof. – Rynant May 31 '15 at 18:30
  • @Tyrth My post with the solution for 2 of each shape got deleted because it wasn't really an answer to the question. Would you like to include it in your post, so we can keep all the interesting info on this puzzle together? That'd be great. – Sumyrda - remember Monica Jul 27 '15 at 20:15