A VERY HARD RIDDLE NOOO ONE CAN SOLVE!!!!!!!!!

I don't suppose I am right as it only took 2 mins

Originally posted by stevetodd
Think I have guessed it? You get on the cales with all bags, drop (dispose of a bag) and if the scale drops by an amount less than (is it 10 grams sorry forgot what the weight was) the weight of the real coins then the last bag dropped is the bag with fake coins. Is that the correct answer?
Steve

I thought of that too, but unless that is part of the riddle, I decided that would make than one "weighing" as you are actualy weighing 10 times (worst case) to find the bag.

Originally posted by Faith No More
thats correct zaps! if for example the fake coins weigh 5 grams and the weight in the end is lets say 535 grams so you know that bag 3 is fake

shaul

well, then you would be knowing how much the fake coins weigh.

so you must say the weight of the fake coins are known.

Other wise at 535 grams total you could think that bag 5 is fake, if the fake coins are 3 grams each.

or bag 6 is fake, if fake coins are 2.5g

;-) so we must have known how much the fake coins weigh to get the answer in one weighing.

Actually you can do it without weighing anything, get ten friends and give some coins to each friend, each friend having coins from a different bag. Tell them to go to the shops and buy something with the money and bring it back to you. Anybody who didn't make it back should have been arrested for using counterfeit money. Then it should be easy to deduce which of the bag(s) had the false money in. 😛

Originally posted by jaredboudreau
well, then you would be knowing how much the fake coins weigh.

so you must say the weight of the fake coins are known.

Other wise at 535 grams total you could think that bag 5 is fake, if the fake coins are 3 grams each.

or bag 6 is fake, if fake coins are 2.5g

;-) so we must have known how much the fake coins weigh to get the answer in one weighing.

you are right sorry about that. i dint tell this riddle in a long time so i forgot some details.

shaul:

Originally posted by Faith No More
you are right sorry about that. i dint tell this riddle in a long time so i forgot some details.

shaul:

You don't need to know exactly how much the fake coins weigh, but if you guarantee that they weigh less than 9 grams, then the trick becomes determining the optimal number of coins required from each bag to confidently determine which one is fake.

For example, if you took 1, 2, 3, 4, ... 10 coins from each bag respectively, and found the total to be two grams light, you wouldn't know whether the fakes were 2 nine gram coins or 1 eight gram coin.

However, taking 1, 10, 100, 1000, ... coins would be sufficient.
Note then that the number of coins in each bag does become important. You need a milliard (billion in American) coins in one of the bags for such an exercise. 😉

Originally posted by craigy
You don't need to know exactly how much the fake coins weigh, but if you guarantee that they weigh less than 9 grams, then the trick becomes determining the optimal number of coins required from each bag to confidently determine which one is fake.

For example, if you took 1, 2, 3, 4, ... 10 coins from each bag respectively, and found the total to be two g ...[text shortened]... ant. You need a milliard (billion in American) coins in one of the bags for such an exercise. 😉

Ok folks, what is the minimum weighings if you had one pair of balance type scales, 10 bags of coins, and knew only that 1 bag contained fake coins which had a different (higher or lower) weight to the real coins?

I made this up just now and I'll be thinking up the answer at the same time as you lot - let's see who wins!

Originally posted by iamatiger
Ok folks, what is the minimum weighings if you had one pair of balance type scales, 10 bags of coins, and knew only that 1 bag contained fake coins which had a different (higher or lower) weight to the real coins?

I made this up just now and I'll be thinking up the answer at the same time as you lot - let's see who wins!

Ah, but this reduces to a problem we've had before. IIRC the answer is 3.

Prove it then Mr drunken Shogun... Tell us the answer!!!!!!!!😠😛

Originally posted by mortalmatt
Prove it then Mr drunken Shogun... Tell us the answer!!!!!!!!😠😛

It's a little convoluted, but quite possible.
You only need 1 coin from each bag.

1)
Separate your coins into three groups a(3 coins), b(3 coins), c(4 coins).
Weigh a and b.

A: If a and b balance then all coins in a and b are genuine, and the fake is in c. (This is the easy case)
A2) Take 2 genuine coins from a or b and weigh with two coins from c.
This narrows down which two coins from c contains the fake.
A3) Weigh a genuine coin against one of the pair that includes the fake, and presto - you'll have your solution.

B: If a and b don't balance, things get a little more complicated, because now the fake is one of 6 (we have no clue as to whether the fake is in a or in b).
To resolve this you need to take note as to which of a and b was lighter. (Without loss of generality, we'll assume a was lighter than b).

B2) Now take two coins from a (we'll call this a1) and two coins from b (we'll call this b1). Weigh this with the 4 genuine coins from c.

C: If the sets balance then your fake is one of the two coins left behind.
C3) Weigh one of those two coins with any of the known genuine coins and you can identify the fake.

D: If the sets don't balance, then the fake is in a1 or b1. (This is where the fact that a was lighter than b becomes important.)
If a1 + b1 is lighter than c, then the fake coins are lighter than the genuine ones and the fake is in a1. Otherwise the fake is in b1.
Either way you have reduced identifying the fake to one of 2 coins.
D3) Simply requires weighing one of the possible fakes with a genuine coin as the final step.