Blue and Red Marbles

Logic Level 3

A bag contains 80 blue marbles and 70 red marbles. Next to this bag is a pile of blue and red marbles. We take the marbles out of the bag in groups of two, with the following rules:

  • If the two marbles are of the same color, we put them in the pile and place a blue marble back into the bag.
  • If the two marbles are of different colors, we put them in the pile and place a red marble back into the bag.

What is the color of the last marble left in the bag?

Blue Red It can be either color

Laszlo Mihaly by Laszlo Mihaly , 57, USA

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1 solution

Laszlo Mihaly
Oct 10, 2018

At each step the number of red marbles is either unchanged, or reduced by two. Therefore if we started with an even number of red marbles, the number of red marbles will always be even.

The game ends when we have two balls left, and we pull these two balls out. It is impossible for one of them to be blue and the other one red, because in that case the number of red balls would be odd. Therefore in the last step we either pull two blues or two reds and we replace them with one blue. That is the last ball left in the bag.

8 Helpful 0 Interesting 2 Brilliant 0 Confused

Can you give a bit more detailed explanation?

si cheng - 2 years, 8 months ago

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What part of the solution are you interested in? The main point is that in the last step we put back a red marble ONLY if we took out one red and one blue. But that is impossible, since the number of red marbles must be an even number.

Laszlo Mihaly - 2 years, 8 months ago

Although I think the problem is pretty much justifiable, I think it will look better if you add the "game-ending" condition. I made mistake and thought that the game would end if only one color remains in the bag. The question itself implies that the game will end after only 1 ball is left, but it still leaves slightly a bad taste as it's somewhat ambiguous

Afkar Aulia - 2 years, 5 months ago

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Can you suggest a wording?

Laszlo Mihaly - 2 years, 5 months ago

Your solution is great. With other words: After the last taking we take back one marble, so only one marble can left. Assume that the last marble is red( R R ). We can replace an R R by R B RB and a B B by R R RR or B B BB . The parity of the amount of R is constant( R → R B = 1 → 1 ; B → R R / B B = 0 → 0 / 2 R\rightarrow RB=1\rightarrow 1;B\rightarrow RR/BB=0\rightarrow0/2 ). Therefore the last marble can't be R R . The solution is B \boxed{B} .

A Former Brilliant Member - 11 months, 4 weeks ago

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