EuroBillTracker

You have 12 sacs of rice..

11 have same weight , one is different (a little more or less , to you unknown)

You can weigh 3 times on an oldfashioned balance-weight.

How can you find out which sac has a different weight?

Enjoy the riddle

EDITED - Incorrect answer (should teach me to read things more thoroughly...)

(BTW, you could've put that in this thread)

the answer is incorrect.

You assume that one is heavier , but in the riddle it says you don't know if one is heavier or lighter..

So ; Try again..

You take 1 sac and put that on 1 side of the oldfashioned balance-weight.

Than you put another sac at the other side, if the oldfashioned balance-weight is in balance than you know that 1 of the sacs have the same weight

You leave 1 sac on the oldfashioned balance-weight at the right side. Than you put 1 by 1 al the sacs at the left side and if the oldfashioned balance-weight not in balance you know that that sac at the left side is the sac with the difference

You can only weigh three times, that would take 11 times...

First I also thought it was as BogPoet said because someone told me a riddle like this, but it said that one was havier than the others. I don't know the answer for this one.

It is not an easy one , but it CAN be done..

First you weigh 3 sacs on each side. If they weigh the same you can put them aside and turn to the other 6. If they don't, you put the other 6 sacs aside.

Then you weigh one group of the first 3 sacs and one group of the new 3 sacs. If they weigh the same, you can put them aside. If not, you can put the last group of 3 sacs aside.

Now you have 3 sacs and one time to weigh... You know now that one of them is lighter or heavier.

So, you put on one side 1 of the sacs that you know is "normal" and 1 of the new ones and on the other side the same thing. If they weigh the same, the different one is the one that is left. But if they don't weigh the same... I'll bang my head on the wall!

I can't solve it, GRRRRRRR, my head is letting smoke out...

nice try though

I have another one ;

You have five sacs with in each sac 10 coins.
One of the 5 sacs contains 10 false coins. The other 4 sacs have only real coins.
A real coin is always 10 gram , a false coin 11 gram.
You have a real weight-scale , and you are allowed to weigh 1 time.
How do you find out in which sac the 10 false coins are??

I really don't know.

But what about solving the first one first? And sorry if I'm wrong, but isn't this a Brain teaser?

FVS wrote:nice try though

I have another one ;

You have five sacs with in each sac 10 coins.
One of the 5 sacs contains 10 false coins. The other 4 sacs have only real coins.
A real coin is always 10 gram , a false coin 11 gram.
You have a real weight-scale , and you are allowed to weigh 1 time.
How do you find out in which sac the 10 false coins are??

easy one :

take one coin from the first bag, two coins from the second, three from the thrid etc.
If all coins were real, you would have 150 grams. The number of grams higher indicates how many false coins there are (e.g. 152 grams means there are two false coins, so bag two contains the false coins)

for the other riddle, i really don't know

Titania's answer was the thing I tried first, but it turned out to be insufficient. It would only work if you knew whether the different sac is heavier or lighter, just like BogPoet's procedure. But BogPoets suggestion was more interesting. While Titania divided the sacs succesively into two groups of equal size, BogPoet divided them into three groups, and he did not treat them equally in the first step. As a result, in some cases BogPoet was able to find the heavier sac (if it had been known that the sac was indeed heavier) in just two steps, whereas Titania always needed to weigh three times. Thus let us review his procedure.

BogPoet divided the twelve sacs into three groups of four. One group was put aside, the other two were put onto the left and the right scale, respectively. If these weigh the same, we know that the different sac is in the third group of four, and by weighing twice we could find the different one (check it!). But if the eight sacs do not have the same weight, it seems that we only know that one of them has a different weight. But this is not true, we know more: we know which side is heavier and which one is lighter, and we know which sacs are on the left scale and which are on the right scale. And we know that the four sacs that have been put aside must have the same weight. We may use some sacs that we have put aside in the first step, we may put some other sacs aside, we may leave some sacs on their scale and we may transfer some sacs from one scale to the other one.

Good luck to all of you!

PS: If I have not made a mistake, you can find out in eleven out of twelve cases, whether the different sac is heavier or lighter than the other ones.

@Venga ; correct , I hadn't noticed that topic

@Jules ; you solved it very nice.

the first riddle is still waiting to be solved.. it is a difficult one , that's true..

FVS wrote:the first riddle is still waiting to be solved.. it is a difficult one , that's true..

It's a nice riddle!
I have kind of a solution which works in most cases but not in all cases. Maybe someone can work out a better solution (and maybe a better way to describe it).

(1)
Put 6 on the scale, 3 left and 3 right.
If the scale is in balance, the 6 on the scale are the reference group (all have the same weight). Name them R. The others contain the fake sac, name them F.
If the scale is out of balance, the 6 which are not on the scale are the reference group (all have the same weight). Name them R. The ones on the scale contain the fake (but you don't know on which side). Name them F.

(2)
Take 3 from R and put them on the left side of the scale.
Take 3 from F and put them on the right side of the scale. Name them X.
Name the remaining ones of F to Y.
If the scale is in balance, Y contains the fake.
If the scale is out of balance, X contains the fake, and you know whether the fake is heavier or lighter!

(3)
You now have a group of three which contains the fake.
Put 2 of them on the scale, one on either side.
If the scale is in balance, the remaining one is the fake.
If the scale is out of balance, and the group was X, the fake is on the scale and you know on which side it is.
If the scale is out of balance, and the group was Y, and the scale was out of balance in (1), the fake is on the scale and you know on which side it is.
If the scale is out of balance, and the group was Y, and the scale was in balance in (1), the fake is on the scale but you don't know on which side

micro is goin g in the right direction ; but as he mentioned himself ; it is not 100%..

Next one??