Twelve coins problem

[pdurrant]

A variant on a classic problem.



I have twelve apparently identical coins. I keep them safe in a small bag, tucked away under my jacket. Well, in fact in two bags — so that even if one bag gets stolen, I won't have lost all my money (eggs and baskets, you know). The bags are not very pretty — just mass-produced things — but they are strong enough to hold the coins securely.

I suspect that one of my coins is a fake, and is either slightly heavier or slightly lighter than the others, and I need nine good coins to buy something I need to get today. Passing fake money is severely punished, so I can't just take a chance by picking nine of my twelve at random.

Unfortunately, the only scales available to me are owned by a miser who charges one coin per weighing — and I can't risk paying him with a fake coin either!

His scales are simple balance scales, two pans in which coins can be placed, and either the pans balance, or one side drops while the other rises.

Oh — and no-one's willing to lend me a known good coin or coins to weigh against my coins. A very unfriendly place, this.

Finally, I don't even have a pencil or paper. I'm going to have to work this out without writing anything down...

How can I end up with nine good coins from my twelve, and at the same time manage to pay the miser with good coins for however many times I need to use his scales? No matter what I do, I obviously can't afford to use his scales more than twice...





If you post a solution, don't forget spoiler tags. Thanks!

[omk3]

Pdurrant, I'm enjoying your puzzles very much

Spoiler:

Let's see now.
I have twelve coins, so I can divide them into 4 groups of 3, A, B, C and D.

I put A and B on the scales.
If they don't have the same weight, the fake coin is in either A or B.
If they do, the fake coin is in either C or D.

If the fake coin is in either A or B, I weigh A with C.
If they have the same weight, the fake coin is in B.
If they don't have the same weight, the fake coin is in A.

If the fake coin is in either C or D, I weigh A with C again.
If they have the same weight, the fake coin is in D.
If they don't have the same weight, the fake coin is in C.

I have now identified a group of 3 coins, one of which is the fake one. I give all three to the miser, and he can work out for himself which one is the fake one and throw it away, he still gets 2 good coins for my use of the scales.

[pdurrant]

Originally Posted by omk3

Pdurrant, I'm enjoying your puzzles very much

Yes, an excellent solution. You're good at this!

[Catire]

Oops never mind I didn't read it before answering

[omk3]

Originally Posted by Catire

or

Spoiler:

Split the coins into 3 groups of 4 coins (A,B and C)

First you weigh A and B, if they have the same weight then the coin is in group C, if either of them weight more or less than the other then the coin is in that group. Discard the lighter or heavier coin groups.

You then split the group where the coin is into two groups of two and weigh them, wichever is heavier or lighter thats where the coin is. Discard the heavier or lighter group.

Then you just weigh the two coins left and you've found the fake one.

But:

Spoiler:

First, you used the scales more than two times.
Second, you say

and weigh them, wichever is heavier or lighter thats where the coin is.

Obviously one is heavier, one is lighter. Which one has the fake coin? You don't know if the fake coin is lighter or heavier.

(Edit) Sorry, I answered before you deleted your solution...

[Catire]

Originally Posted by omk3

(Edit) Sorry, I answered before you deleted your solution...

I immediatly thought of the "classic" one without bothering to read carefully

[pdurrant]

Originally Posted by Catire

I immediatly thought of the "classic" one without bothering to read carefully

Yes, slightly different constraints.

[obs20]

Okay here it goes:

Spoiler:

We need 9 good coins and 2 coins to pay the skinflint for the scale. We shouldn't get upset to offer him 3 coins of which 2 are good. Let him figure it out himself.

Put 3 coins on the left side of the scale.
Put 3 coins on the right side of the scale.
If the scale balances great we know that we have 6 good coins.
Then
Take 3 coins off the left side and replace them with 3 of the remaining coins.
If the scale balances great we have 9 good coins.
If not we know that the 3 remaining coins are good and we can use those.

If the scale didn't balance in the first place we still take 3 coins off the left side and replace them with three new coins. If the scale balances then all the coins on the scale are good and also the 3 remaining coins are good adding to 9.

If the scale still doesn't balance then we know that 1 of the 3 coins on the left are bad and we can use all the others to pay 9.
In either case we only uses the scale twice.

[obs20]

Originally Posted by omk3

Pdurrant, I'm enjoying your puzzles very much

Spoiler:

Let's see now.
I have twelve coins, so I can divide them into 4 groups of 3, A, B, C and D.

I put A and B on the scales.
If they don't have the same weight, the fake coin is in either A or B.
If they do, the fake coin is in either C or D.

If the fake coin is in either A or B, I weigh A with C.
If they have the same weight, the fake coin is in B.
If they don't have the same weight, the fake coin is in A.

If the fake coin is in either C or D, I weigh A with C again.
If they have the same weight, the fake coin is in D.
If they don't have the same weight, the fake coin is in C.

I have now identified a group of 3 coins, one of which is the fake one. I give all three to the miser, and he can work out for himself which one is the fake one and throw it away, he still gets 2 good coins for my use of the scales.

Way to go!

[dreams]

Here goes:

Spoiler:

Lt scale 5 - Rt scale 5 -- off to the Rt side 2

Scale is even so the fake is off to the side

Put those on scale to find the fake.

Pay miser 2 coins, toss the fake coin, and you keep 9 coins.

or if uneven..

L scale 5 - R scale 5 -- off to the R side 2

L scale is light, so put R scale bag to the side with other 2 coins

Then take one coin out of the L bag and put off to the Lt side

Now, L scale 2 - R scale 2 -- off to the Lt side 1

Scale is even so the coin off to the Lt side is fake

Pay the miser 2 coins, toss the fake coin, and you keep 9 coins.

or if uneven

L scale 2 - R scale 2 -- off to the Lt side 1

L side is light

Pay the miser 1 good coin from the Rt side and ask the miser if he wants the light L bag and he can weigh it himself to find the fake one?

[pdurrant]

Originally Posted by obs20

Okay here it goes:

Yes, that's a good solution.

[pdurrant]

Originally Posted by dreams

Here goes:

Alas, you've missed the bit in the question that says "either slightly heavier or slightly lighter than the others" and have assumed that a fake coin will always be lighter than a real coin.

Try again? You're looking at it along the right sort of lines. And yes, it's acceptable to pay a debt of two coins with two good coins and one fake one.

[dreams]

And again

Spoiler:

Divide the coins into 4 bags. 3 coins in each bag =12

1- Weigh any 2 bags. If the same weight then remove one and add the 3rd bag to the scale.

2- If the 3rd bag:

• weighs the same, then the fake is in the 4th bag. Pay the miser with the 4th bag.

****

1-Weigh any 2 bags. If uneven weight then remove either the lighter or heavier bag and add the 3rd bag to the scale.

2- If the 3rd bag:

• weighs the same, then the fake is in the bag that was just removed. Pay the miser with that bag.

• results in uneven, remove the bag that matches the weight of the first one removed (lighter or heavier) . The fake is in the bag still on the scale.

I give up. I have to be up in 4 1/2 hours and I am brain dead.

[pdurrant]

Originally Posted by dreams

And again

[solution]

I give up. I have to be up in 4 1/2 hours and I am brain dead.

Brain dead, but RIGHT! Congratulations.

[pdurrant]

Originally Posted by Catire

I immediatly thought of the "classic" one without bothering to read carefully

You can still have another go — unless you've already checked one of the correct answers.