Comment: Really beautiful problem

This solution is wrong. I didn't see another possibility of the situation (a very common mistake in Combi problems).Ragib Farhat Hasan wrote: ↑Mon Nov 04, 2019 9:32 pmThe problem can be approached by first cutting the board in half; let the upper-half be called $BROWN$ and the lower-half be called $GREEN$.
Now, a pair of brown worms each from the top-left corner move to each of the squares (except the last one) in the diagonal of the LHS $1007\times1007$ square of the $BROWN$ region after a finite set of moves. Only one brown worm occupies the last square of the diagonal.
Now, it can be implied that when each two worms in the mentioned squares each moves to the right and down directions respectively, with none of the worms' paths intersecting, all the unit squares in the entire $BROWN$ region is occupied at least once.
The same procedure can be carried out in the $GREEN$ region, only this time the worms will move right and up.
Therefore, the minimum total number of worms $=> 1006\times2\times2+2=4026$.
Because there is no active member!Ragib Farhat Hasan wrote: ↑Tue Nov 05, 2019 12:57 amBTW, I wonder...
This problem was posted over 5 years ago.
And no one posted a solution to this problem in half a decade!!! WOW!!!