The surface of the 9 × 9 × 9 cube is covered completely with 1 × 2 pieces of paper. Some of those pieces cover two squares on the same face, but some bend along cube’s edge. Prove that the number of bending pieces is odd.
This seems pretty similar to a regular 2d case, so let's make every square either white or black, however this wouldn't work at a corner, so I have no idea how to tile. Thanks!
You'd rather tile the "small cubes".
There are $9 \times 9 \times 9$ small cubes, each of size $1 \times 1\times 1$. Paint them black and white, alternately, so that all $8$ corners are painted black.
Now notice that a bending piece covers two squares of the same color, while a non-bending piece covers two squares of different colors.
It only remains to calculate the number of black and white squares. This is the same on each face, which is $41$ black and $40$ white. Therefore in total there are $6$ more black squares than white squares.
Let $a$ (resp. $b$) be the number of bending pieces covering two black (resp. white) squares. Then $2a - 2b$ is the difference between black and white squares, which is $6$.
So $a - b = 3$ is odd, hence $a + b$ is also odd.