a) Since in any operation, black ants move to white and white and moves to black, at least one black square remains empty.

b) In the 9×9 square, peel layer by layer. Each layet other than centre has an even length and can have a clockwise movement of bug. The centre but moved to step 2 and shares the place with some other bug thus making one spot empty but others in the right spot. Proof by construction.

]]>For the final sum to be zero, all the numbers will belong to one of the two groups – group 1 or group 2 such that ( sum (group 1)) – (sum (group 2)) = 0

For example, lets the number be 6,7,8,9 then 6 and 9 will belong to group 1 and 7 and 8 to group 2.

Now, sum(group 1 & 2) = 2* sum (group 1) which implies sum(group 1 & 2) = even number

Sum of (1,2…., 1989) is an odd number, therefore it is not possible to get 0 finally. ]]>

Operation applied on 1 and 2 give 1 and 1. And then applied on 1 and 1 gives 0 and 0.

Now for any number x,

Operation applied on 0 and x give x and x. And then applied on the two numbers gives 0 and 0.

Apply the operation on all the numbers and get zeroes. Done.

Solution seems too simple though. May be I understood the question wrong?

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