TITLE: Number of rectangles with odd area.
QUESTION [8 upvotes]: We have a $10\times 10$ square. 
  How many rectangles with odd area are on the picture? 
 
  I say lets choose a vertex first, there are $11\cdot11=121$ possibilities. 
  Now, choose odd width and side (left or right) and odd length and side (up or down). There are $5$ possibilities to each, so in total we have $\dfrac{121\cdot 25}{4}$ rectangles. We divide by $4$ because every rectangle is counted $4$ times, one time for each vertex of the rectangle. But, the result is not a whole number. Where am I wrong? Thanks.

REPLY [3 votes]: I assume you're looking for rectangles with sides parallel to those of the square and vertices that are lattice points.  A rectangle with integer sides has odd area iff the sides are both odd.  For each pair of odd integers $x, y$ with $1 \le x,y \le 9$, there are $(11-x)(11-y)$ possible positions for a rectangle of size $x \times y$.  Thus the number of rectangles is
$$ \sum_{x \in \{1,3,5,7,9\}} \sum_{y \in \{1,3,5,7,9\}} (11-x)(11-y) = 900$$