-- Posted by steiny at 4:05 pm on Dec. 14, 2006 There are 10 stacks of coins, each stack has 10 coins in it. One stack is counterfeit.  The real stacks weigh 100 grams, thus each coin weighs 10 grams, the counterfeit stack weighs 90 grams, thus each coin weighs 9 grams.  You have a scale, that measures weight to the gram, you may only use it once. How do you figure out which stack is counterfeit? -- Posted by FurryPanther at 4:07 pm on Dec. 14, 2006 Add one stack at a time. When the scale changes only 90 grams, you've found the stack. FP -- Posted by dovelove at 4:07 pm on Dec. 14, 2006 Hold each stack of coins in your hands and weigh them against one another. Whichever feels the lightest, put it on the scale. -- Posted by BsRdSxs12 at 4:08 pm on Dec. 14, 2006 Put all of the stacks on the scale, then take them off one by one. When the weight goes down by 90 grams instead of 100, you have figured it out. -- Posted by dovelove at 4:08 pm on Dec. 14, 2006 Quote: from FurryPanther at 7:07 pm on Dec. 14, 2006 Add one stack at a time. When the scale changes only 90 grams, you've found the stack. FP Oh. That's better than my answer. =( -- Posted by MachinegunHead at 4:11 pm on Dec. 14, 2006 You split all the stacks into different ratios of themselves, i.e. stack one into 10 coin pile, and one 0 coin pile, stack two into a 9 coin pile and a 1 coin pile, etc. 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. -- Posted by steiny at 5:08 pm on Dec. 14, 2006 so far all wrong, its like an electronic scale so it shows the total gram weight on the scale, and u only get one measurement so you can't add or take away -- Posted by MachinegunHead at 5:37 pm on Dec. 14, 2006 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. -- Posted by steiny at 5:44 pm on Dec. 14, 2006 Quote: from MachinegunHead at 8:37 pm on Dec. 14, 2006 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. on the right track but its not a two sided scale, u put something on it and it shows the weight -- Posted by MachinegunHead at 5:47 pm on Dec. 14, 2006 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. -- Posted by steiny at 6:08 pm on Dec. 14, 2006 Quote: from MachinegunHead at 8:47 pm on Dec. 14, 2006 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. Solution * * * * * * * * * * * 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. -- Posted by MachinegunHead at 6:23 pm on Dec. 14, 2006 Yeah, haha, that is basically what I was trying to say, just hadn't thought enough about it to condense it into a paragraph. I figured I'd just go about the quesiton mathematically and hope my effort becomes fruitful.