## Puzzle #180: White and Black Beans

This is a wonderful puzzle for all ages from Alok Mittal’s weekly Math Circles class. The puzzle is doable by everyone, so please do try yourself and also have your children try this one.

A pot contains 75 white beans and 150 black ones. Next to the pot is a large pile of black beans. A somewhat demented cook removes the beans from the pot, one at a time, according to the following strange rule: He removes two beans from the pot at random. If at least one of the beans is black, he places it on the bean-pile and drops the other bean, no matter what color, back in the pot. If both beans are white, on the other hand, he discards both of them and removes one black bean from the pile and drops it in the pot. At each turn of this procedure, the pot has one less bean in it. Eventually, just one bean is left in the pot. What color is it?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy bean counting!

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### 4 Responses to Puzzle #180: White and Black Beans

1. Praneeth says:

White will be the last one remaining!
Reason:- At any time there will be odd number of whites remaining.

2. Vishal Poddar says:

White. White can be discarded only in a pair. Initial number is odd, so finally when 1 bean is left, it will be white.

3. Prakhar says:

Alok – thanks to the Mathematical Study Circle Book which you had once recommended, my son Anahat had seen similar problems like this one before. So he answered this question as White bean remaining because bean initial is odd for white and changes in white happen in even terms. So odd – even will be odd ; so the last remaining bean will be white

4. The parity of white never changes in any operation. So white cannot become zero from 75. So the last bean has to be white bean.