662) No.
1) Every living person has shaken hands with a certain number of other persons . prove that the count of the number of the people who have shaken hands an odd number of time must yield an even number.
2) In a chess is it possible for a knight to go from the lower left corner of the board to the upper right hand corner in the process to light exactly once on each square ?
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10 Answers
661) If one looks at the total number of handshakes completed at any moment, we see that this number must be even. On the other hand, the total number of handshakes is also the sum of the handshakes made by each person. So the number of persons who have shaken hands an odd number of times must be even (since odd times even = even).
341The condition of having to alight on each square once seems redundant to me as the knight will never land on the diagonally opposite square. If you label the initial square as (0,0), then the destination is (7,7).
Notice that no matter which square, with coordinates (x,y), it lands x+y is invariant mod 3. Here initially it is 0 mod 3 and now its obvious that it cannot touch the last square.
Anything wrong here?
3Sir , as i think , here its all dependent on how the knight moves . if it starts from black colored square then in next step it will fall on white and if one step again then it will lie on black again . even number of moves means the knight will be on the same colored square where as with an odd number of moves it will be on different colored square . in this case its on black colored square and the destination is also black square and hence with an odd number of moves (63) it will never move on to the destination square .
62Prophet sir, I think it should not be x+y modulo 3 is constant
bcos the knight can have either x+1, y+2 or also.. x+1, y-2
that too is an allowed move.. so x+y will not be constant mod 3..
3@ prophet sir i found that if the knight starts from 0,0 then at every move x+y ≡ 0 mod 3 satisfies and hence it will never move to ( 7,7)
662) I thought it out this way. In order to traverse the chessboard, while stopping precisely once on each of the squares, the knight must move 63 times. At each move the knight goes from a square of one color to a square of another color. Thus, after an even number of moves the knight is again on a square of the same color that it started from, and in an odd number of moves it is on a square of the other color. Therefore, the knight cannot arrive at the opposite end of the diagonal of the chessboard in 63 moves since the initial and the final squares are of the same color.
62I second the solution by anant sir.
I guess a simpler solution would be that
x+y ≡ n-1 mod 2
where n is the move number and x, ;y are the coordinates at the nth move..
We start with the 1st move.... *(I mean the initial state is called to be the first move)
I din write that bcos I felt that minch had effectively said the same thing [1]
341Unfortunately for me I had first thought of the same soln, but somehow in my mind the knight landed in the same colour square :(