1. ## Pairs and pieces

Find all such pairs of positive integers (m,n), for which a rectangle with dimensions of m x n can be built from such pieces created from 6 squares marked as "O" (note that you can both rotate and flip the pieces):
Code:
O O
OOOO
(if it's not clear, it's a F-type shape)

OK. So we now that we can create a 3x4 rectangle by simply putting together this shape and the one rotated by 180^:
Code:
OOOO
O O
As we can create 3x4, we can also create 4a x 3a for a positive integer a. But what else? I'm kind of clueless...

2. ## Re: Pairs and pieces

Hello, GGPaltrow!

Find all such pairs of positive integers $\displaystyle (m,n)$,
for which a rectangle with dimensions $\displaystyle m\!\times\!n$ can be built
from pieces created from 6 squares as shown.
(Note that you can both rotate and flip the pieces):

. . $\displaystyle \begin{array}{cc}\square \;\,\square \;\; \\ [-2mm]\square\! \square\! \square\! \square \end{array}$

OK, so we know that we can create a 3x4 rectangle
. . by simply putting together this shape and the one rotated by $\displaystyle 180^o.$

. . $\displaystyle \begin{array}{c}\blacksquare\! \blacksquare\!\blacksquare\! \blacksquare \\ [-2mm] \square \! \blacksquare\! \square\! \blacksquare \\ [-2mm] \square \! \square \! \square \! \square \end{array}$

We can also create $\displaystyle 4a\times 3b$ rectangles for positive integers $\displaystyle a$ and $\displaystyle b.$
But what else? .I'm kind of clueless ...

Me too . . .

I've tried to find other possible rectangles, but have failed so far.

3. ## Re: Pairs and pieces

Hmm.. maybe sth like this:

Suppose we have:
Code:
O O
OOOO
in which the four squares form an edge. Then, let's try to put sth beside it (as putting above leads to a 3x4 rectangle):

Code:
   OO
O
O OOO
OOOOO
This can't be transformed into a rectangle as our "F" can't be put inside it in any way. The same goes with this flipped:
Code:
    OO
O
O O OO
OOOOO
Which is even more wrong as we leave our black F laying and blocked from the "open" side so we can only complete it to a 4x3.

Making any F lie on its "back" beside the black one doesn't make sense cause it leads to completing each of them to a 4x3 which means we eveantully end with 4ax3b.