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You are given ten stacks of golden coins, each stack consisting of ten coins and a digital scale with arbitrary precision.

You know that all stacks of coins are made from gold, weighing 10 grams per coin except one stack, which is made from silver but is painted golden. A silver coin weighs 1 gram less than a golden one, which makes its weight 9 grams.

With only being able to weigh exactly once, how can you identify the stack containing the silver ones?

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    $\begingroup$ Now solve the problem if any number of the stacks could be of silver coins, but each stack has 1,000 coins. $\endgroup$ – Joe Z. Jul 30 '14 at 1:53
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    $\begingroup$ @JoeZ. use powers of 2, instead of incrementing by 1 $\endgroup$ – Joe Oct 9 '14 at 0:08
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  1. You can number stacks from 1 to 10.
  2. From each stack take as many coins as the stack number.
  3. Weigh them all.
  4. Subtract the weight from 10+20+...+100=550. The result is the number of the silver stack.

P.S. You can do the same even if you have 11 stacks of 10: just number them from 0 to 10.

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Get a yard stick with a known weight. Attach one end to a fulcrum point, and rest the other end on the scale. Place the stacks of coins at equal distances along the yard stick.

Next calculate what the scale should read if all ten stacks contained ten gold coins. Then take a reading from the scale. This will come up slightly less then what was calculated. Based on that difference, calculate the position along the yard stick where the lighter stack must be.

Note: I read this wrong the first time. I assumed that the silver stack contained one silver coin among 9 gold. The above solution will work in either case.

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