Vaggelis has a box that contains $2015$ white and $2015$ black balls. In every step, he follows the procedure below: He choses randomly two balls from the box. If they are both blacks, he paints one white and he keeps it in the box, and throw the other one out of the box. If they are both white, he keeps one in the box and throws the other out. If they are one white and one black, he throws the white out, and keeps the black in the box. He continues this procedure, until three balls remain in the box. He then looks inside and he sees that there are balls of both colors. How many white balls does he see then, and how many black?
Problem
Source: Greece JBMO TST 2016 p4
Tags: combinatorics, game, game strategy
EVTrainer
29.04.2019 05:08
The only way the number of black balls within the box decreases is if two black balls are chosen, in which the number of black balls decreases by 2. Therefore, the number of black balls always remains odd and there are 1 black and 2 white balls at the end
Viktor_Maks
19.11.2022 20:55
1) Both balls are black * Because one is painted white and the other one is thrown out of the box, the number of black balls gets smaller for 2. 2) Both balls are white * The number of white balls in the box gets smaller for two. 3) One is white and one is black Only the number of white balls is reduced by 1. From $1)$, $2)$ and $3)$, we can see that the parity of the black balls in the box is always odd. Because at the end $1 \leq$ $black$ $balls$ $< 3$ => There is only 1 black ball at the end. The number of white balls is $3 - 1 = 2$. Our conclusion is that at the end, there are: * $2$ white balls * $1$ black ball
huashiliao2020
27.03.2023 23:04
The invariant is the parity of black balls, which is odd. So there are 2 white balls and 1 black balls left at the end.
arfekete
27.03.2023 23:11
Literally just mod 2 Either white decreases by 1, or white increases by 1 and black decreases by 2. So black is always odd.