# Solving the Fake Coin Problem: Find It in Three Weighings

• T@P
In summary: If the scale balances: Weigh: 1&2 vs 4&3 Otherwise the lighter side contains both phoniesIn summary, this conversation is about a scheme to determine if a coin is fake by weighing it three times and checking to see if it balances. The first time, if the coin balances, then it is a fake coin. If it does not balance, the lighter side contains both phonies.
T@P
there are a lot of these fake coin/find it in three weighings etc problems. i have a few that i like, and maybe someone here can add some that they like/know? just an idea :)

anyway this is a funny one, if you have 6 coins and 2 of them are fake, the question is can you determine (in just *3* weighings) which coins they are assuming they are lighter than the other 4 real coins?

i have trouble spelling many things

oh and the scale tells you only which side is heavier. (duh)

Code:
[COLOR=black]
Number the coins 1-6
Weigh: 1&2 vs 3&4
If scale balances:
Weigh: 1&3 vs 2&4
If scale balances:
Weigh: 1&4 vs 2&3
If scale balances: 5 & 6 are the phonies
otherwise the lighter side contains both phonies
otherwise the lighter side contains both phonies
otherwise, name the lighter-side coins L1 and L2.
Weigh: L1 vs 5
If scale balances:
Weigh: L1 vs 6 (scale cannot balance)
If L1 is lighter, then L1 and 5 are the phonies
If 6 is lighter than L2 and 6 are the phonies
If 5 is lighter, it is a phony and L2 is the other phony.
If L1 is lighter:
Weigh: L1 vs L2
If the scale balances, L1 and L2 are the phonies
Otherwise 6 and L1 are the phonies
[/color]

That's interesting and i believe correct but wouldn't it be easier to split the coins into two halves. Three and three, if they balance weigh two from each half and from that you can figure out which coins are fake, if the originals don't balance then you only have to weigh once more to determine the fake coins.

That's interesting and i believe correct but wouldn't it be easier to split the coins into two halves. Three and three, if they balance weigh two from each half and from that you can figure out which coins are fake, if the originals don't balance then you only have to weigh once more to determine the fake coins.
I don't think that would work. After they balance the first time and you weigh 2 from each half, say that the 2 balance. Now either the lighter coins are both contained in the two pairs you weighed, or both are not. You can't determine where they are with only the one remaining weighing.

if they balance a second time then the coin which wasn't weighed is the lighter one and you can move on to weighing the other half.

Because there are only two fake coins if they balance the first time it means there is a fake coin in each half. If it doesn't it means that the lighter side contains both fake coins, hence you only have to weigh one more time after that.

Here's an interesting question which I do not know the answer to. If you have n coins total and r fake ones out of those n, how many weighings do you need to separate the fakes coins from the genuine coins?

For that question, that's assuming that you can only weigh equal #'s of coins. i.e. no determining the weight of a fake coin by trying, say, 4 known-real coins vs 5 known-fake coins.

I'm guessing there are many different solutions to the same question depending on how many fake ones there are and if they are an odd or even number of them. There is an algorithm that works for every combination (weighing half and half until you eliminate each possibility), but to get an efficient process you'd need to know n and r. Therefore, depending on what process you use and luck you may have to weigh many times or fewer in certain cases. In short i don't know either.

what said:
if they balance a second time then the coin which wasn't weighed is the lighter one and you can move on to weighing the other half.
There were 2 that were not weighed in the second weighing in your scheme.
Code:
Weigh 1&2&3 vs 4&5&6
If the scale balances
Weigh 1&2 vs 4&5
If the scale balances
You're out of luck.

Code:
Weigh first half(1&2&3) vs second half(4&5&6)
If the scale balances
For first half:
Weigh 1 vs 3
If the scale balances
2 is lighter.
otherwise you have found either 1 or 3 to be fake.
For second half:
Weigh 4 vs 6
If the scale balances
5 is lighter.
otherwise you have found either 4 or 6 to be fake.
If the scales don't balance set aside the lighter side three coins name them 1,2,3
Weigh 1 vs 3
If the scale balances
they are the fake coins.
otherwise you have found either 1 or 3 to be fake and 2 is fake by elimination.

Okay, that works. Your original description was unclear.

by the way, is fake coins 'always' lighter.
because i once given this kind of puzzle by a professor and he use stone instead of coins. and he also added that the weight is 'diffrent' without mentioning wether the 'fake' is lighter of heavier. n i it still ringing in my head until now.

Do you remember how many fakes and reals there were and how many weighs you had?

wait a second: whay if you have say n coins, and all n of them are fake. how on Earth are you going to find the n fake coins if you don't know how many there are (!)

theres also a funny one that's slightly different,

you have 10 sacks of coins. each sack contains *many* coins. you know that one sack is entirely fake, and that the difference in weight between that sack and the others is exactly 1 unit. in one weighing using a scale that tells you the exact difference in weight determine the bag containing the fake coins.

this one is funny in that the last part is a little confusing and can be done in the easy and harder way(s)

I've heard this one before.

Not too hard.

Weigh all the coins. If the scale tips both the light coins are on the same side, and all you have to do is eliminate the heavy one. You can do this in one more weighing because if they don't balance you know which is heavier and if they do balance the one you didn't weigh is heavier.

If the scale balances, do the one weighing for each side to find out which are the lightest.

Easy stuff!

Alkatran said:
Not too hard.

Weigh all the coins. If the scale tips both the light coins are on the same side, and all you have to do is eliminate the heavy one. You can do this in one more weighing because if they don't balance you know which is heavier and if they do balance the one you didn't weigh is heavier.

If the scale balances, do the one weighing for each side to find out which are the lightest.

Easy stuff!
No, the question is to find the counterfeit bag in one weighing.

sorry I've been sortof delinquent on this post. anyway i didnt hear any answers to it and it has been awhile so ill just say the solution:

you take 1 coin from bag 1, 2 from bag 2, 3 from 3 etc and put them on one side of the scale. then take 45 (sum of 1 - 10) from *any* bag and put it on the other side. clearly if the 45 same coins are real than the weight difference corresponds to the number of the bag with fake coins. if theyre fake you'l get a number that much bigger than 10 and you should know that the 45 are fake and either know which bag you took them from or go through some easy math to calculate which one it is.

sadly i can't recal a really hard one, maybe someone here knows it...? its something to do with 2 weighings and like 10 coins and one is fake (lighter)... anyone?

hope I am not borring anyone here, i think its cool personaly...

T@P said:
sadly i can't recal a really hard one, maybe someone here knows it...? its something to do with 2 weighings and like 10 coins and one is fake (lighter)... anyone?

The closest I can come is: 9 coins, 1 is light, find the light one in 2 weighings on a balance scale. But it isn't really hard. The problem scales in the sense that you can start with 27 coins and find the light one in 3 weighings, etc.

yes it goes by factors of three mostly, since if you can show the fake one is in a pile of 3 you are done, so it goes by factors of three... maybe it was only hard because i knew it like 5 years ago...

well anyone know any other ones?

if i had not mistaken, 12 stones, 3 times weighing, and 1 stones of diffent weight.

actually i just remembered a really funny one, it goes like this:

6 coins, 2 fake coins, heavier by 1 gram. in 4 weighings find the fake coins *but* the scale you are using will only register a weight difference of at least 2 kg. think of it as a regular balance scale with friction. (somewhat more realistic really)

its actually quite hard too (at least for me)

oh and the one with 12 stones is kind of easy. first weighing is 6 - 6, then you know the fake one is in a group of six. second is 3-3 from the group of 6, then last weighing is 1-1 from the group of 3 with fake coin. btw I am assuming you *know* the coin is lighter (or heavier)

Last edited:
yes, if you know.
but he say that we don't know the weight

## 1. What is the "Fake Coin Problem"?

The Fake Coin Problem is a mathematical puzzle that involves identifying a counterfeit coin among a group of identical looking coins using only a balance scale and three weighings.

## 2. How does the "Fake Coin Problem" relate to real-life situations?

The Fake Coin Problem is often used as a thought exercise to demonstrate logical reasoning and problem-solving skills. It can also be applied to real-life situations where identifying a counterfeit item is important, such as in the banking industry or in counterfeit detection.

## 3. What is the best strategy for solving the "Fake Coin Problem"?

The most efficient strategy for solving the Fake Coin Problem is to divide the coins into three groups, weigh two of the groups against each other, and then use the results to determine which group contains the fake coin. This process is repeated until the fake coin is identified.

## 4. Are there any variations of the "Fake Coin Problem"?

Yes, there are several variations of the Fake Coin Problem, including ones with a different number of coins, different types of fake coins, or using a different number of weighings. These variations can make the problem more challenging and require different strategies to solve.

## 5. Why is the "Fake Coin Problem" important in the field of mathematics?

The Fake Coin Problem is important in the field of mathematics because it demonstrates the use of logical reasoning, problem-solving skills, and strategic thinking. It is also a fun and engaging way to introduce mathematical concepts to students and challenge their critical thinking abilities.

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