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/