Every ”corner” covers exactly 3 squares, so necessary condition that tilling exists is 3|mn. First, we shall prove that for tilling with our condition it is necessary that both m, n for m, n > 3 must be even. Suppose contrary, that m > 3 is odd (without losing generality). Look at ”corners” that cover squares on side of length m of table m × n. Because m is odd, there must be ”corner” which covers exactly one square on side. But any placement of that corner forces existence of 2 × 3 rectangle in tilling. Thus, m and n for m, n > 3 must be even and at least one among them is divisible by 3. Notice that in corners of table m × n ”corner” must be placed as in figure. If one of m and n is 2 then condition forces that only table is 2 × 3 or 3 × 2. If we try to find desired tilling when m = 4 then we are forced to stop at table 4 × 6 because of the conditions of problem. We easily find example of desired tilling for table 6 × 6 and tilling for 6 × 2k. Thus, it will be helpful to prove that desired tilling exists for tables 6k × 4l, for k, l ≥ 2. Divide that table at rectangle 6×4 and tile that rectangle as we described. Now, change placement of problematic ”corners” as in figure. Thus, we get desired tilling for this type of table. Similarly, we prove existence in case 6k × (4k + 2) where m, l ≥ 2. But, we first divide table at two tables 6k×6 and 6k×4(l−1). Divide them at rectangles 6×6 and 6×4. Tile them as we described earlier, and arrange problematic ”corners” as in previous case. So, 2 × 3, 3 × 2, 6 × 2k, 2k × 6k ≥ 2, and 6k × 4l for k, l ≥ 2 and 6k × (4l + 2) for k, l ≥ 2.

Where is the figures?