TargetIIT

8 pieces are placed on a chessboard such that no two pieces lie on the same row or column.

Prove that an even number of pieces lie on black squares

[b]Since there are only 4 black boxes in a period, we can see that if the piece opting for a white box (let's call it x) goes to an even period, no black box remains to be filled in the 7th period - i.e before we reach upto the 7th period, all the 4 black boxes are exhausted. X takes away 1 black box, while the remaining black boxes are exhausted by the given rule by the pieces in the odd periods.
Same thing happens for P-8 when x takes a seat in the even periods.

For (3w,5b) - the same logic follows.
If all the 3 occupy odd (or even periods), P-8 (or P-7) has no black box left .
Same logic if majority (i.e. 2) occupy odd or even periods....as before to prove that black boxes in P-8 and P-7 are exhausted before they are reached....[i][/i][/b]


not even a single line goes into my head... can u please use some simpler english and explain wat u want to say..

i would be thankful to u if u do dat

Let us number the columns and the rows as 0,1,2...8.
For the ith row and the jth column to be white, we must have (i+j) as even, and conversely for a box to be black, (i+j) must be odd.

Now whatever be the way we put these pieces on the chess board, the sumtotal of i's or j's will be some permutation of the sum of the 1st 8 natural nos. Total Σ(i+j)=72, which is even.

If we try to input 5 pieces on white squares and 3 on black squares, then we must have an eqn of the form

2a+2b+2c+2d+2e+(2k+1)+(2m+1)+(2n+1)=72, which is not satisfied by any natural soln.

Similalrly if we try to put 1 piece in a black box, we have 2k=71, which again has no natural soln.

This proves that odd number of pieces cannot lie on black squares.

yup, that's it!