Vangelis has a box containing 2015 white and 2015 black balls. He follows the following procedure: He chooses randomly two balls from the box. If both are black, then paints one of them white and puts it in the box, while drops the other out of the box. If both are white, then he keeps one of them in the box and drops the other out of the box. If one ball is white and the other is black, then he keeps the black in the box and drops out the white ball. He follows the procedure, until only 3 balls remain in the box. Then he realizes that in the box exist balls of both colors.
Determine how many white and how many black balls exist finally in the box.
Solution
We will consider what happens at every step of the procedure for the number of white and black balls. In the case of selection of two black balls, no one is going back to the box and so the number of the black balls is reducing by 2. In the second case of selection of two white balls, the number of the black balls has not any change. In the third case of selection of balls of both colors, also the number of the black balls in the box remains invariant. Therefore we observe that, in every step, the number of the black balls in the box remains invariant or it is reducing by 2. Since their initial number is 2015 (odd) at every step their number will remain odd. Therefore in the case of 3 balls in the box of both colors, we conclude that one of them must be black and the other two must be white.
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