FP

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I think we are in rats' alley
Where the dead men lost their bones.

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Add one stack at a time. When the scale changes only 90 grams, you've found the stack.

FP

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Oh. That's better than my answer. =(

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when i play doctor, i play to win

then you put one half of the piles on one side of the scales and the other half on the other side. So 10 of stack one on the left, 0 on the right, 9 of stack 2 on the left, 1 on the right, etc.

so then going by the ratio at which the weights are out by to figure out exactly which stack it is. As the imbalance will be exact for a certain configuration of the counterfeit coins.

Am I right?

To people above: He did say use them only once. Taking things off and reweighing is breaking the rules of the puzzle.

Post edited at 4:16 pm on Dec. 14, 2006 by MachinegunHead

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A four course orgy at the end of your bed.

Just say stack 7 is the counterfeit one, and you split all stacks the way I proposed, then you'd have:

Original stack: left/right  (ratio of the coins in the relative stack)

1: 100 / 0 (10:0)
2: 90 / 10 (9:1)
3: 80 / 20 (4:1)
4: 70 / 30 (7:3)
5: 60 / 40 (3:2)
6: 50 / 50 (1:1)
7: 36 / 54 (2:3)
8: 30 / 70 (3:7)
9: 20 / 80 (1:4)
10: 10 / 90 (1:9)

so USING THE SCALES total left weight = 546g
and THROUGH ASSUMPTION total right weight = 444g

then you compare this with the expected outcome (Assuming they are no counterfit ones)

which is:
left weight = 550g
right weight = 450g

So then you can just work backwards.

so change in left is 4, change in right is 6.

so the ratio of the change in left to the change in right is 2 : 3.

Working down the original ratio scale (which is the same regardless of whether a stack is counterfit or not) to see if there is a ratio of 2/3 gives us that the counterfit stack is no.7

So yes, I AM right, thank you very much.

The assumption is allowed seeing as you specified that there is only ONE illegitimate stack.

edit: I edited it just to make it clear that I only used the scale once.

Post edited at 5:42 pm on Dec. 14, 2006 by MachinegunHead

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A four course orgy at the end of your bed.

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I think not steiny, my idea works perfectly.

Just say stack 7 is the counterfeit one, and you split all stacks the way I proposed, then you'd have:

Original stack: left/right  (ratio of the coins in the relative stack)

1: 100 / 0 (10:0)
2: 90 / 10 (9:1)
3: 80 / 20 (4:1)
4: 70 / 30 (7:3)
5: 60 / 40 (3:2)
6: 50 / 50 (1:1)
7: 36 / 54 (2:3)
8: 30 / 70 (3:7)
9: 20 / 80 (1:4)
10: 10 / 90 (1:9)

so USING THE SCALES total left weight = 546g
and THROUGH ASSUMPTION total right weight = 444g

then you compare this with the expected outcome (Assuming they are no counterfit ones)

which is:
left weight = 550g
right weight = 450g

So then you can just work backwards.

so change in left is 4, change in right is 6.

so the ratio of the change in left to the change in right is 2 : 3.

Working down the original ratio scale (which is the same regardless of whether a stack is counterfit or not) to see if there is a ratio of 2/3 gives us that the counterfit stack is no.7

So yes, I AM right, thank you very much.

The assumption is allowed seeing as you specified that there is only ONE illegitimate stack.

edit: I edited it just to make it clear that I only used the scale once.

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on the right track but its not a two sided scale, u put something on it and it shows the weight

Post edited at 5:48 pm on Dec. 14, 2006 by MachinegunHead

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A four course orgy at the end of your bed.

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I know.... I put the left one on it, and because you know the total weight (including the counterfit coins) is 990, then to figure out the weight of the right hand side you just do 990 - 546, which is 444g.

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Solution
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so yea ure pretty much right.  You just could have worded it simpler.  Say you take 1 from the first pile, 2 from the 2nd, 3 from the 3rd, 4 from the 4th, 5 from the 5th, 6 from the 6th, 7 from the 7th, 8 from the 8th, 9 from the 9th, and 10 from the 10th. Assuming they were all real it would weigh 550 grams. If it is one gram less (549 grams) it's the first pile, 2 grams less (548 grams) it's the second pile... 10 grams less (540 grams) it's the 10th pile.