- 27 Jan '05 10:46 / 1 edit
You have 10 sacs of coins. All the sacs contain coins that weigh the same = 10 grams. One sac of coins is odd and contains coins that weigh 9 grams. How to find out which sac contains the odd coins with only one try?

You have 10 sacs of coins. All the sacs contain coins that weigh the same = 10 grams. One sac of coins is odd and contains coins that weigh 9 grams. How to find out which sac contains the odd coins? - 27 Jan '05 13:04

put 1 coin from the first sac, 2 from the second,..., 10 from the tenth.*Originally posted by yevgenip***You have 10 sacs of coins. All the sacs contain coins that weigh the same = 10 grams. One sac of coins is odd and contains coins that weigh 9 grams. How to find out which sac contains the odd coins?**

if all coins weighed the same this selection should weigh 55.

If it weighs 54 it's the first, 53 it's the second...45 is the tenth. - 28 Jan '05 04:52

It works if you are allowed to use standard weights and measures , along with the balance. If separate standard weights are not permitted then it cannot be done in a single " balancing" act.*Originally posted by genius***ignore me - Palynka's idea works better and takes only one weigh...** - 28 Jan '05 12:38

well-if you just have the scales you could weigh 5 bags against 5 and remove one bag from each side until both sides balance. when this happens, the bag you just removed from the formally lgihter side is the one your looking for! i thought it was quite an ingenius method, but then looked at playnka's worked better etc etc etc...but i still like mine cause i thought of it and meh!*Originally posted by CoolPlayer***It works if you are allowed to use standard weights and measures , along with the balance. If separate standard weights are not permitted then it cannot be done in a single " balancing" act.** - 28 Jan '05 15:40

The original problem as stated did not mention any standard weights involved.*Originally posted by yevgenip***That's right!**

Another solution could very well have been that you take two more balances, and weigh once on each of them. But that's simply cheating

Be more specific in your questions! - 28 Jan '05 17:26

That's smart. A further improvement ,in terms of simplification, would be taking 10 coins from each of the 10 bags and place the coins from 5 bags on one side and those from the other 5 bags on the other side of the scale. Thereafter the procedure of collectively removing the coins of one bag at a time, could be done till balance , as in your procedure.*Originally posted by genius***well-if you just have the scales you could weigh 5 bags against 5 and remove one bag from each side until both sides balance. when this happens, the bag you just removed from the formally lgihter side is the one your looking for! i thought it was quite an ingenius method, but then looked at playnka's worked better etc etc etc...but i still like mine cause i thought of it and meh!**