The awkward bag
You have 17 blue balls and 17 red balls in a bag. You want to free yourself of them, but you have one condition. You have to take out 2 ball at once. If you take balls with different colors, you will put another red ball. If you take balls with same color, you will put another blue ball. What will be the color of the last ball in the bag?
Important: No matter what, each time you take out 2 balls, you have to put another one(the correct one). Even if you someway get 2 balls left, and you take out these 2 balls, you have to put another ball.
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1 solution
It's impossible to get free from all red balls. Each time u take 2 red balls, you shall add a blue ball. You had an odd number of red balls and you still have an odd number. If you take out 2 blue balls, you'll still have an odd number for red balls. If you take out 1 blue and 1 red ball, you'll lose 1 red ball, but you have to add another one, so won't change at all.
You will always have an odd number of red balls. In the end, no matter how many blue balls you have, you shall have 1 red ball. If you take out this red ball plus one blue, u must return the red ball. If you take out 2 blue balls, you will have to return one blue ball(if you still have 2 or more blue balls).It's impossible take out 2 red balls if you just have one.
When you get 1 red ball left, it's impossible to get free of it. No matter what, each move you do( 2 different colors or the same), you would just get away from 1 blue ball.