Finding the Bad Machine by Gumball Weight
Date: 05/13/2008 at 17:48:06 From: Greg Subject: One bad gum ball machine. You have ten gum ball machines. Nine make 1 ounce gum balls and 1 makes gum balls that look exactly the same but only weigh .5 ounce. You do not know which machine is the defective one. You are able to use a scale one time to figure out which machine is making the half ounce gum balls. How do you solve? I heard it on the radio, and it's driving me nuts! I can't see how the weight of the gum balls have anything to do with figuring out which machine is making the lighter ones? It has to be quantity related. If you weigh 10 gum balls they should equal 9.5 ounces.
Date: 05/13/2008 at 18:20:32 From: Doctor Achilles Subject: Re: One bad gum ball machine. Hi Greg, Thanks for writing to Dr. Math. That's a tough one. The challenge is that you only get to use the scale once. Let's try solving a simpler problem: You have 2 gum ball machines. One is defective and produces 0.5 ounce gum balls, the other works fine and produces 1 ounce gum balls. How can you find the defective machine with only one weighing on one scale? This problem isn't so bad. You just label the two machines "A" and "B", then take one gum ball out of machine A. If it weighs 1 ounce, then machine B must be broken, but if it weighs 0.5 ounces, then machine A is broken. BUT: THERE IS ANOTHER WAY TO SOLVE THIS SIMPLE PROBLEM! Instead of taking one gum ball out of machine "A" and zero gum balls out of machine "B", what if you took one gum ball out of machine "A" and TWO gum balls out of machine "B" and then put them on the scale TOGETHER? Well, there are two possible outcomes: I) One gum ball from A + 2 gum balls from B could weigh 2.0 ounces. In that case, you can deduce that machine B was defective. Do you see why? II) One gum ball from A + 2 gum balls from B could weigh 2.5 ounces. In that case, you could deduce that machine A was defective. Again, do you see why? Now, here is my challenge for you. Can you extend this logic to three machines? How about up to the ten machines in your problem? I hope this helps. If you have other questions or you're still stuck, please write back. - Doctor Achilles, The Math Forum http://mathforum.org/dr.math/
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