There's an well-known puzzle that goes as follows: You have 12 balls. They look identical, and 11 of them have the same weight but one of them, you don't know which, has a different weight. You don't even know whether the odd ball is heavier or lighter than the others. You have an old-fashioned balance scale (they look like this if you're not familiar with them), and you can put balls on the two sides of the scale and weigh them against each other. The scale is really rusty, so it will break after using it three times. So how can you use the three weighings to determine which of the 12 balls has a different weight than the others?(adsbygoogle = window.adsbygoogle || []).push({});

You might want to try your hand at it, but I have just thought of the following twist on the traditional puzzle: Suppose that you don't have a regular balance scale that tells you which side is heavier and which side is lighter, but instead you had some kind of crippled balance scale that would only tell you whether the two sides are equal or unequal, not which side has a greater weight. Given this modification, how many weighings would suffice to solve the problem, and how would you solve it? I'm guessing that 4 or 5 weighings would suffice, but I'm not sure.

I look forward to hearing people's thoughts on this.

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# 12 Ball Puzzle with a Twist

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