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Imagine a piece of paper. It has a square grid of 1x1 on it, so that every square has an area of 1 cm(squared). That piece of paper was folded into the shape of an (empty, hollow) cylinder whose length is 50 cm and whose base circumference is also 50 cm (look at the picture below). Can you cover the area of that cylinder with the shape on picture b, which is made up of 4 squares and is also of dimensions 1x1? Sorry for the bad picture!

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    $\begingroup$ What have you tried so far? Also where did you get this from, this question seems very familiar. $\endgroup$ – Benson Lin Oct 16 '16 at 12:25
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    $\begingroup$ Have you at least a faint idea of the answer (possible/not possible ?). $\endgroup$ – Jean Marie Oct 16 '16 at 12:39
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It is impossible :

The number of squares in cylinder is $50^2$

And we color black or white in them, like chess board

Hence at block in b) we have two coloring ways : 3 black and 1 white, 1 black and 3 white If the number of blocks of first type is $x$ and the number of second types is $y$, then $$ 3x+y=50^2/2 $$ $$ x+3y=50^2/2 $$ so that $ x+y=625$

I will receive JeanMarie's suggestion and I will use TonyK's argument : $x+y=625$ That is the number of 4-square-tiles is odd Hence WLOG we can assume that $x$ is odd and $y$ is even Hence the number of black squares in cylinder is $$50^2/2=3x+y$$ Right hand side is even and left hand side is odd. Hence it is a contradiction.

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    $\begingroup$ Very good solution [+1] to a problem that could have been asked on a plane checkerboard. $\endgroup$ – Jean Marie Oct 16 '16 at 13:16
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    $\begingroup$ $4x+4y = 50^2 \implies x+y = 25^2$, not $25^2/2$. $\endgroup$ – Joffan Oct 16 '16 at 13:36
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    $\begingroup$ More simply: the number of tiles is $50^2/4$, which is odd. Each tile has $1$ or $3$ black squares. So the number of black squares is odd. But there are $50^2/2$ black squares, and $50^2/2$ is even. $\endgroup$ – TonyK Oct 16 '16 at 13:54
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    $\begingroup$ @HK Lee I suggest you to add, at the end of your text, something like "which is equal to 625, thus an odd number. Contradiction." $\endgroup$ – Jean Marie Oct 16 '16 at 14:48
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    $\begingroup$ A different way to prove it is to realize that in order to have an equal number of black and white squares in the end, there must be an equal number of tiles with each coloring. But a pair of tiles of different coloring has 8 squares in total, so the total number of squares must be divisible by 8. $\endgroup$ – kasperd Oct 16 '16 at 22:35

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