A tiling puzzle/question My teacher gave us a riddle that goes like this:
You have a $7\times 7$ square and $16$ $3\times 1$ tiles.
Of the $16$ tiles, $15$ are straight and $1$ is crocked ("L" shaped).
When you tile the square with these tiles you should get that one unit is left un-tiled (because $7 \times 7=49$ and $16\times (3\times1)=48$).
The question is in what locations can the un-tiled square be?
Keep in mind that you can rotate the pieces. 
I've never saw this kind of questions before so I'm not sure how to go about solving something like this. I tried to check some positions and it seems that the "L" shaped tile cannot be placed in the corners, but I don't know how to continue...
Any help will be appreciated because this is driving me crazy.
Thanks.
 A: Here is a basic 4×4 arrangement of 4 straight tiles and one crooked tile:
| | r--
| | | |
| | X |
----- |

Divide the 7×7 square into four blocks of size 4×4, 4×3, 3×4, and 3×3. The latter three are easily tiled by 11 straight tiles. Rotations of the above arrangement in the 4×4 block yields four different possible positions of the hole. Thanks to Gerry Myerson's answer, these are the only possible positions, up to rotation of the whole square.
A: This was written in response to the old version of the question, which asked where the L could go and remarking that it could not go in a corner.  The coloring is still valid for the new on about where the 1x1 goes.
As Rahul Narain points out, a first line of attack is coloring.  If you number the columns $0$ to $6$ and the rows $0-6$ as well you can put a number in each square by summing the row and column modulo $3$.  This is like coloring a chess board with three colors.  You have $17$ squares marked $0$ and $16$ with $1$ and $2$.  Each straight tromino covers one square of each number, but the L may not.  If the L covers two $1$ cells or $2$ cells you are sunk.  This has to be true for a given placement no matter how you rotate or reflect the numbers.  This prevents you from putting the corner of the L in a corner, but it doesn't prevent you from putting one end in a corner, and there is a solution:
ABBBCCC
AADDDXE
FFFGGGE
HHHIIIE
JJJKKKL
MMMNNNL
OOOPPPL

where the X is the empty square.
A full solution would show what other positions are allowed and how they are possible.
A: I think this coloring works. Color each row 0 1 2 0 1 2 0 (so the 1st column is all 0, the 2nd is all 1, etc.). This gives you 21 zeros, 14 ones, 14 twos. A straight tile covers three the same (if it's vertical) or one of each (if it's horizontal). If we say it covers $a$ zeros, $b$ ones, and $c$ twos, then in either case $a\equiv b\equiv c$, modulo 3. It follows that the four spaces left after you place the 15 straight tiles are 4 0 0 (that is, 4 zeros, no ones, no twos), or 2 1 1 or 0 2 2. Now the L has to cover some permutation of 2 1 0, and the uncovered square has to be some permutation of 1 0 0. Some of these work, e.g., 2 1 0 plus 0 0 1 gives 2 1 1. But there is no way to add 1 0 0 to any permutation of 2 1 0 to get any of 4 0 0, 2 1 1, or 0 2 2, so the uncovered square can't be a zero, that is, it can't be in the 1st, 4th, or 7th column. 
Applying the transpose, it can't be in the 1st, 4th, or 7th row, either. That leaves only 12 places where it can be. By symmetries of the square, these 12 places are of only three types, so if you can find a tiling leaving each of these three types uncovered, you're done. Ross has found a tiling leaving what I'd call $(2,2)$ uncovered; now it remains to do $(2,3)$ and $(3,3)$. 
