c4countdown

Here is a question somebody decided to set on a worksheet I had to do.

You have a set of balancing scales, and 12 £1 coins. You are told that one of the coins may be a forgery, and if it is then it will have a different weight than all of the others. It is always possible with 3 weighings to work out if there is a forgery, and if so, if it's heavy or light.

See if you can devise a system that works.

JackHurst wrote:Here is a question somebody decided to set on a worksheet I had to do.

You have a set of balancing scales, and 12 £1 coins. You are told that one of the coins may be a forgery, and if it is then it will have a different weight than all of the others. It is always possible with 3 weighings to work out if there is a forgery, and if so, if it's heavy or light.

See if you can devise a system that works.

I remember seeing this on Dara O'Briain's School of Hard Knocks on Dave.

This is the correct answer if I remember correctly:

Split the weights into 3 groups of 4. Put one group in one scale and another group in the other. If both weigh the same, then the other group has the forgery.

Then in that group, split it into 2 two's and then split the lighter or heavier group and and whichever is heavier/lighter is the forgery.

James Robinson wrote: Split the weights into 3 groups of 4. Put one group in one scale and another group in the other. If both weigh the same, then the other group has the forgery.

But what if they don't?

The second step isnt right. With your method, you could successfully work out there are no forgeries, but it could be the case that you have ended up with 2 coins, one of which is heavier than the other, but you dont know if its because one is a light forgery, or one is a heavy forgery. Nice try though.

I like this http://laughmaths.blogspot.co.uk/2011/0 ... try-2.html

This is a great question, I can't get my head around how to solve it if we aren't told in advance whether the forgery would make the coin lighter or heavier.

Edit:
I thought we had to isolate the exact coin that was the forgery. That would be a cool step to the question, might take what, 5 or so moves?

Answer:
Split them up into 3 groups of 4, then weigh off AvB, AvC, and BvC, one of the pairings will balance out (let's say AvB balances), so the forgery is in C.
To know if the forgery is heavier or lighter then just analyse the results of AvC and BvC (either C will be heavier than both A and B, or C will be lighter than both A and B).

I'm always suspicious of exact wordings in this kind of question, and the wording here is ringing alarm bells - it doesn't specify that you have to pick a particular coin out of the pile and say "This one is fake".

Problem is, I can answer the question as asked in only two weighings. Weigh six against six. If they balance, there is no forgery. Otherwise weigh the light pile three against three; if they balance, there's a heavy forgery in the other pile; if not, there's a light forgery on the scales; either way, the asked question is answered.

Do we need to isolate the forgery?

Nick Deller wrote:I'm always suspicious of exact wordings in this kind of question, and the wording here is ringing alarm bells - it doesn't specify that you have to pick a particular coin out of the pile and say "This one is fake".

Problem is, I can answer the question as asked in only two weighings. Weigh six against six. If they balance, there is no forgery. Otherwise weigh the light pile three against three; if they balance, there's a heavy forgery in the other pile; if not, there's a light forgery on the scales; either way, the asked question is answered.

Do we need to isolate the forgery?

Awesome job sir

http://www.mathplayground.com/coinweighing.html

I reckon, you can isolate the coin in 3, otherwise surely this website wouldn't exist

Andy Platt wrote:http://www.mathplayground.com/coinweighing.html

I reckon, you can isolate the coin in 3, otherwise surely this website wouldn't exist

Yes, it's possible to correctly isolate the heavier or lighter coin in 3 weighings. Great problem that I originally saw many years ago.
Keep thinking!

I thought this was going to be about that game when you ask a girl if you can weigh her tits and then motorboat her as you go "wee---ayyy!!". But it's not.

Matt Morrison wrote:I thought this was going to be about that game when you ask a girl if you can weigh her tits and then motorboat her as you go "wee---ayyy!!". But it's not.

I think I'd prefer this one

Yeah sorry, just to clarify, you are supposed to be able to isolate the coin and say if it is heavy or light.

If any of you are still stuck, I can give you a hint:

Coins that have been weighed and for which the scales were balanced can be used again as a control. So say you have narrowed it down to {coin A is heavy} or {coin B is light}, then doing AB vs CD will give you the answer

I got that hint very quickly myself but it still didn't help me

People still doing this? The thing about just knowing whether the forgery is heavy or light and not knowing which one - you can do that in two steps whether you have 12 coins or 12 million. Anyway, I haven't read the clue, and I might still have another look at this.

I don't think you can possibly do this in two steps, having seen the three step answer.

New weighing puzzle I saw somewhere:

You have a balance scale and six balls. Two are red, two are green, and two are blue. You are told that, in each pair of like colored balls, one is slightly heavier than the other. The heavy balls in each pair are all the same weight, as are the light balls. Using two weighings, develop a system that always allows you to pick out the three heavy balls.

Steve Balog wrote:You have a balance scale and six balls. Two are red, two are green, and two are blue. You are told that, in each pair of like colored balls, one is slightly heavier than the other. The heavy balls in each pair are all the same weight, as are the light balls. Using two weighings, develop a system that always allows you to pick out the three heavy balls.

I'd seen the puzzle in the OP a long time ago (and then seen the solution, thinking "Christ, I would never have got that") but this one looks a lot more simple. But I still can't do it. As far as I can see, the first weighing pretty much has to be one of the following:

a) RBG v RBG - whichever side is heavier contains at least two of the heavy balls
b) RRG v BBG - whichever side is heavier contains the heavy G

But I can't see a second weighing for either

Hint for my puzzle: The first move in my puzzle is sort of unintuitive to consider at first.

Think I got the one in the OP after like a half hour:

Weigh four coins against four other coins in the first weighing.

If the coins balance, all 8 are legit. Weigh three unknown coins against three legit coins. If they are heavy, the forgery is heavy and is on the scale. If they are light, the forgery is light and on the scale. Either way, weigh two of the suspected forgeries against each other. If they balance, the straggler is the forgery, and if not, whichever side you are looking for (light or heavy) is the forgery. If that second weighing balances, weigh the last coin against any legit coin. If it's heavy, the last coin is a heavy forgery, if light, it's a light forgery, and if they balance, there's no forgery.

If the coins do not balance, the four coins not weighed are legit. Weigh three coins on the heavy side and two coins on the light side against the four known legit coins and the fourth possibly heavy coin. If they balance, then weigh the two coins not on the scale against each other. If one is light, it's a light forgery If they balance, there is no forgery. If the side with the three heavy and two light coins in the second weighing is heavy, then one of the the heavy coins from the first weighing is the forgery, and it's heavy. Weigh any two of the three suspects against each other. If they balance, the third is the forgery, if not the heavy one is. If the side with the single heavy coin is heavy, then either the heavy coin on that side is heavy, or one of the two light coins on the other side is light. Weight the heavy candidate and one light candidate against two legit coins. If the unknowns are light, the light side coin is a light forgery. If they are heavy, the heavy coin is the forgery. If they balance, that light coin you didn't weight is a light forgery.

I think this is it.