Anyone solved it?

https://brilliant.org/100day/day0/

Thanks Here is one of the solutions for 67×100: To end in a pocket, the ball must travel an integer number of table widths and an integer number of table heights at the same time. Since 67 and 100 are coprime, the first time this happens is when the ball has travelled 67 table widths and 100 table heights. At this point, the ball will have crossed the table horizontally an odd number of times, and so will be at the right-hand edge. It will also have crossed the table vertically an even number of times, and so will be at the bottom edge. Thus it will be in the bottom-right corner, and go into pocket D. For 2013 and 1000, notice that (2013,1000)=1 and 2013 is odd and 1000 is even and then we continue analogously as above

The cordinates of A is (0,0), of B is (0,2013) of C is (1000,2013) and of D is (1000, 0). It is easy to prove that every time the ball hits the edges, is will be in points with cordinates with the same parity (for example (2k, 2l) or (2k+1, 2l+1)). So since B has cordinates with different parity, it is obvious that the ball will never reach to it!