$$A(0, 0)$$, $$B(4012, 0)$$, $$C(4012, 6018)$$, $$D(0, 6018)$$ determine a rectangle which is the union of $$4012 \times 6018$$ unit squares. Determine the number of these unit squares which have a nonempty intersection with $$\overline{AC}$$.

The diagonal $$\overline{AC}$$ has slope $$\frac{6018}{4012} = \frac{3}{2}$$ and $$\gcd(4012, 6018) = 2006$$, so it splits into 2006 equal pieces, each the diagonal of a $$2 \times 3$$ block, joined at the lattice points $$(2k, 3k)$$. Inside one $$2 \times 3$$ block the diagonal passes through $$2 + 3 - 1 = 4$$ unit squares, so it passes through the interiors of $$2006 \times 4 = 8024$$ squares. In addition, at each of the 2005 interior lattice points $$(2k, 3k)$$ with $$1 \le k \le 2005$$, the diagonal touches two more squares (the upper-left and lower-right ones) at a corner only, and these also have nonempty intersection with $$\overline{AC}$$. Hence the number of squares is $$8024 + 2 \times 2005 = 12034$$.