You have twelve bags filled each with 100 golden coins but one of these bags contains counterfeited coins; good coin weights 5 grams counterfeited one 4.9 gram.. You have a scales with a digital readout, with 5 places of decimals on the display capable of carrying 26 kg on each scale.
How do you determine which bag contains the counterfeited coins? You can use the weighing scale only when paying for it in gold, therefore you get one bag of real gold coins extra. You must pay for using the machine before you use it (it is possible to finance a loan of some sacks: for one borrowed sack you have to pay 1,5 sack back - but then you must bet sure to win the riddle otherwise you'r deep in red). Can you find the counterfeited coins, and how much will it cost you?

Payment for using the scales
(you can put as many sacks on the scale as you want)
using the scales 1x will cost you 1 bag (one try)
using the scales 2x will cost you 7 bags (2 tries)
using the scales 3x will cost you 13 bags (3 tries)

You can keep the gold which you have managed to keep <b>after</b> you pointed out the sack with the counterfeited coins.

Payment for using the scales 1x 1 bag
2x 7 bags
3x 13 bags
doesnt this make it impossible? it costs more to weigh them than they are worth, personally i'd cut my losses and weigh 6 bags at once, if the fakes are in there i'll take the other 6.

The meaning was that you can put as much sacks on the scale as you want. So you can weight all sacks in one try. But I have adjusted the text of the riddle to more clear conditions.

Thing is it costs more bags to use the scales than it is worth, you only have 12 bags without fake coins(including your starting bag), so using any more than that results in a loss, so you can only use the scales twice at most(unless your allowed to use it once for the price of one bag then again for the same price etc).

ok, you lost me.

The riddle is an easy one. Take into the consideration all aspects and you will solve it and go away with lots of sacks....actually and this is a hint you can point the wrong sack by using the scales one times only.

Good luck with it.

OK I make the riddle possibly less confusing:

You have twelve bags filled each with 100 golden coins but one of these bags contains counterfeited coins; good coin weights 5 grams counterfeited one 4.9 gram.. You have a scales with a digital readout, with 5 places of decimals on the display capable of carrying 26 kg on each scale.

You can use the 13th sack if you want (makes the riddle easier because the 13th sack contains surely the gold coins. But you can solve the riddle without using the 13th sack)

You can use the scale how many times you want. Naturally the meaning of the riddle is to use the scale as few times as possible. The person who will use the scale the leats number of times (only once :confused: ) will win this weekend's riddle.

actually and this is a hint you can point the wrong sack by using the scales one times only.
That would be the only solution without losing money, thing is you have 12 sacks, any one could contain the fake coins, if you weigh one bag and are lucky(chances are 12-1 you pick the fake) you find the fake first time, other than that if you weigh 1 and it isnt fake your going to lose money, if you weigh two then one is fake but you cant tell which and the situation is the same for all up to weighing all 12(which would be pointless), if you weigh eleven the one left could be fake(again chances are 12-1 that the fake is left).
But you can solve the riddle without using the 13th sack
Then you'd need a loan to use the scales once, which is pointless.
Im still drawing the conclusion without a large amount of luck it would take up to 4 uses of the scales to find the fake.

so, wouldn't you start out with six bags on each side, then the side that wieghs less you split into three on each side, then you take the side that weighed less, and put two of the bags on the scale. If they weigh the same then the false bag is the one not weighed, otherwise it's the lesser. You get a total of three, right?

so, wouldn't you start out with six bags on each side, then the side that wieghs less you split into three on each side, then you take the side that weighed less, and put two of the bags on the scale. If they weigh the same then the false bag is the one not weighed, otherwise it's the lesser. You get a total of three, right?


3 tries is very good begin, but as I suggested 1 try is enough. A Tip: the scale has 5 decimals precision - what that means....

Look, Tschekhoff (the playwright) said once: in a play everything has to have a function, for example if there hangs a pistol on a wall in a play someone must use it/shoot with it in that play, othewise it disturbs the unity of the play.

The same with the riddle. Why is thus the read out so precise?

Fourty-two!

Easy, I got it in less than a minute... But I seem to be the one always posting the answers to these riddles so I'll just sit here and laugh at you puny humans.

Easy, I got it in less than a minute... But I seem to be the one always posting the answers to these riddles so I'll just sit here and laugh at you puny humans.

Thanks for your (implied) support. I began to worry that the way I put the riddle forward was not OK.

How about you put six bags on each side of the scale, but form bag one you remove one coin, bag two, two coins, bag three three, etc. the side that is short is the side with the fake, and the amount it was off by tells which bag it is.

How about you put six bags on each side of the scale
Good point what type of scales exactly are you using? The type with 2 sides that balance out so you compare the weight or the type where you can only place one object/set of objects to be weighed at a time? Im working on the second type(as thats the only type of digital scales i've seen) is this why im having trouble?

Either way it should work. You just have to do some math.

Either way it should work. You just have to do some math.

Correct. You can use any scale. The trick is in maths. Simple one board scale would do too. Coins differ in weigth. Once you figure the way to identify the sacks (using different nuber of coins from different sacks) it's a piece of cake.

Now how many tries would you need on a two-plate scales if you cannot open the sacks? I came out with 3 tries. Can anyone beat that?

Could it be one if you did it the same way as with one scale. Six on each side, X-1,X-2,X-3, etc. on one side X-1, X-2, etc. on the other.

you put all the sacks on the single scales once, then you start to remove 1 sack at a time and watch the scales... when they show a round number (ending in 00), it means you are holding the counterfeit sack in your hand.

If you are using a pair of scales, then you put 6 sacks on either side and begin to remove 1 sack from each side simultaneously... when the scales show a round figure, you use the scales again to weigh the 2 sacks in your hands and you will find the counterfeit.

But how many times you have to repeat that? I would think that removing the sacks from the scales accounts to a new try.

When I asked the non-open bags question I forget to disable the read out: you can only see if the bags on one scale are heavier then the ones on the other scale:

My thinking was (devide the sacks AAAA BBBB CCCC)

1 try AAAA against BBBB if even then CCCC is devious/lighter
2 try C1 C2 against C3 C4 (eg. C1 C2 side is lighter)
3 try C1 against C2 (solves the riddle)

or if AAAA < > BBBB (eg. AAAA lighter)
then
2 try A1 A2 B3 B4 against B1 B2 A3 A4 (eg. A1 A2 B3 B4 side is EDITTED:LIGHTER (WAS:heavier)
3 try A1 against A2 (riddle is solved)

I think tablariddim has the only possible answer in one weigh otherwise the riddle is impossible to do in one weigh with a single scale. Balancing scales are a different matter.

ok

eg. suppose 6th sack is counterfeid.
one scale:

1 sack 1 coin 5gr.
2 sack 2 coin 10gr.
3 sack 3 coin 15gr.
4 sack 4coin 20gr.
5 sack 5 coin 25gr.
6 sack 6 coin 29.4gr (-0.6 gr.)
7 sack 7 coin 35gr.
8 sack 8 coin 40gr.
9 sack 9 coin 45gr.
10 sack 10 coin 50gr.
11 sack 11 coin 55gr.
12sack 12 coin 60gr.
total: 389.4
(if there were no counterfeited coins you would get 390)
therefore:
390-389.4 = 6 thus 6th sack holds the counterfeited coins

right?

Yes i made a bad assumption that you cannot open the bags and was still working with that.

Yes i made a bad assumption that you cannot open the bags and was still working with that.

That's OK. This weeks puzzle was a kind of chaotic, I will try to put the problem forward more clearly next time.