# Balls of Two Colors II

### Problem

### Solution

The key to solving the problem is to notice that the parity of the number of white balls never changes. Thus the last ball will be white if at the outset the number of white balls was odd; if it was even, the last ball will be black.

Thus, the answer is $1$ if the box originally contained an odd number of white balls and $0$ other wise.

### Acknowledgment

This problem has also appeared in the Outline format that may be more interesting to follow through.

### References

- J. G. McLoughlin et all, Jim Totten's Problems of the Week, World Scientific, 2013, problem #319.

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny