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?
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.