27 Jan 05
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
Originally posted by yevgenipput 1 coin from the first sac, 2 from the second,..., 10 from the tenth.
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
Originally posted by geniusIt 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.😉
ignore me - Palynka's idea works better and takes only one weigh...😛
28 Jan 05
Originally posted by CoolPlayerwell-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!
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
Originally posted by yevgenipThe original problem as stated did not mention any standard weights involved.
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
Originally posted by geniusThat'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.
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!