RAY: This puzzler came to us from Clive Woods. Clive writes:
“I work in the Anti-Counterfeit Department of the U.S. Treasury. The other day, my assistant was sent 100 U.S. quarters, and he found that 10 of them were bo-o-o-o-gus.
“He sorted the 10 bogus coins into one pile, and being an organized sort of chap, he made 9 piles, each containing 10 of the real coins. The weight of the counterfeit coins in this case was different from the weight of a real quarter by 1 gram. However, he forgot whether it was 1 gram more or 1 gram less, but, he knew that the bogus coins were all 1 gram heavier or 1 gram lighter than the real coins.
“He was called away to another job, and he left the 10 piles on his desk. I had to determine which was the bogus pile. To do it, I had a calibrated scale that would tell me the weight placed on it within a fraction of a gram. The question is, how could I figure which was the pile of bogus coins in one weighing?”
Ray: You take one from the first pile, two from the second, three from the third, etc. That gives you a total of 55 coins. Let’s say the genuine coins weigh 10 grams each. A pile of genuine coins should weigh 550 grams. If it weighs 548 grams, the second pile is the bogus pile, and those coins are a gram lighter. If it weighs 552 grams, it’s the second pile, and they’re heavier instead of lighter. Do we have a winner?
Tom: The winner is Russ Bishop from Punxsutawney, Pennsylvania. Congratulations!
(Would someone please explain this to me? I'm not getting it.)
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