The answer is no. (We can't partition the rectangle with an odd number of the regular hexagons.) Here's the proof Assume that there is such rectangle ($x\times y$), then it's easy to see that there exists natural numbers $a,b,c,d$ that $$x=a+b\sqrt{3}, y=c+d\sqrt{3}$$so the area of the rectangle is also an element of $\mathbb{Z}[\sqrt{3}]$. This leads to contradiction since the number of right triangles used is even (This can be easily done with perimeter ), then the area of the rectangle will be $\frac{(2n+3) \sqrt{3}}{2}$.