Weighing in a Harder Way

[Guest]

You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g).

The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter?

I have no answer to this question.

Do you?

[Guest]

yap answer is 4

[Guest]

4

[Guest]

4

I get it in 5

First weigh: 13 13 and 1 not weighed; if 2 are equal then the 1 not weighed is the odd one; otherwise of the 13 which is lighter (or heavier)

Second weigh: 6 6 and 1 not weighed; if 2 are equal then the 1 not weighed is the odd one; otherwise of the 6 which is lighter (or heavier)

Third weigh: 3 3; the 3 which are lighter (or heavier) contains the odd one

Fourth weigh: 1 1 and 1 not weighed; if 2 are equal then the 1 not weighed is the odd one; otherwise the 1 which is not weighed replaces either of those which were (noting if the one left on the scale is lighter or heavier - do not know if the odd one is heavier or lighter do we?)

Fifth weigh: if the one which was replaced is the same as the one removed the other is the odd one; if the two are the same then the 1 removed is the odd one

[Guest]

COINS TO BE DIVIDED INTO THREE GROUPS.

1ST ITERATION (27/3=9 COINS); 2ND ITERATION (9/3=3 COINS); 3RD ITERATION (3 INDIVIDUAL COINS)

MINIMUM MOVES = 3 (IF UR LUCKY)

1)make 3 groups of 9, if on 1st try u weigh equal weights, then 3rd group has odd coin.

2)make 3 groups of 3 coins from the 3rd group i.e from 9 coins which u separated after 1st iteration.

if u lucky and weigh equal weights again, then 3rd group of 3 coins has the odd coin.

3)if u r again lucky enuf to weigh equal coins, then the coin left out is the odd coin.

MAXIMUM MOVES = 4 (EVEN UR NOT SO LUCKY)

1)make 3 groups of 9, if d weights are unequal, then replace one pan with 3rd group,

so it takes 2 TRIES, if in 1st try left pan was heavier and after 2nd try both become equal then u come to know that the group of coins in left pan has the odd coin OR if d left pan is still heavier u obviosly know that is d one having odd coin and also very importantly we can get to know if .... the coin is of 11 g (if pan is heavier)... else of 9g (if pan is lighter) when compared with the other pan.

2)make 3 groups of 3 coins from the 3rd group i.e from 9 coins which u separated after 1st iteration.

now u very clearly know whether the coin is light or heavy ..... so u need not go for 2 tries.

in just one try. if u weigh equal, then 3rd group else which ever is (heavier or lighter) as u already know the weight of odd coin.

so just 1 TRY

3) again just 1 try needed, weigh any 2 coins and can clearly get d odd coin in just 1st try.

THUS ONLY 4 MOVES ARE ENOUGH, IF UR NOT LUCKY ENOUGH TO GET IT IN 3 MOVES.

[Prime]

... You can distinguish among 3**N cases in N weighings. ...

I’ll take up an issue with that statement. I’d say it depends on the kind of possibilities that you are trying to distinguish. The puzzle required to find the counterfeit coin AND tell whether it is heavier or lighter. While you can definitely solve 27 in 4 weighings, I don’t see how you can solve 40 in 4. Although, there are only 80 possibilities (less than 3**4). You can find one counterfeit coin out of 40 in 4 weighings, but not necessarily tell whether it is heavier, or lighter.

For the sake of saving space, solve 13 coins in 3 weighing. I can see how you can find counterfeit coin there, but I can’t ensure determining whether it is heavier, or lighter.

[Prime]

In fact the answer is 2!!!!

You need a minimun of 2 weighs to find the bad coin IF YOU ARE LUCKY!

1- You take 2 coins, you weight them and you get an unbalanced situation.

2- You take one of them and you compare to an outside coin and you still find an unbalanced situstion ==> The bad coin is X and its 9g or 11g

So the minimum ways needed are 2.

First, I’d like to complement the clever redundant use of the factorial sign. (2!!!! = 2).

However, your answer is wrong. The original question was: “… how many minimum ways …”. Whereas, you pointed out only one such way. Indeed, the minimum number of weighings to determine counterfeit coin is 2, if you get lucky. But there are 27!/(25!*2!)=351 ways to compare 2 coins out of 27. And then there are 25*2 ways to compare one of those two to the 25 on the side. Or you could choose to drop 13 coins onto one side of the scale and 13 onto another, and get lucky finding that they weigh the same. And there are C(27,13)*C(14,13) ways, or 27!/(14!*13!)*14 of doing that. Then there are 26 ways to compare the counterfeit coin with one of the true ones. Simplified, the correct answer is: 27! / 24! + 27! * 26 / (13! * 13!). That seems like a very large number, which may not fit into the space allotted by this forum.

(Also, see my previous post for further discussion of the puzzle as it was meant.)

[Prime]

For an in-depth discussion and a maximized version of weighing problem, see my "Ultimate weighing problem" topic on this forum.

[Guest]

Actually 3 weighs are enough.. Its very simple..

[Guest]

First split it into 3 parts of 9-9-9 each.. weigh any two, if they weigh same other part of 9 contains faulty one, else the one which weighs different contains faulty ball..

so

first step--> 9-9-9.

second step-->3-3-3.

thord step--> 1-1-1.

[Guest]

The answer is 4 times. split the pile into two piles of 13 if the scale is even than you have the light coin in your hand. if one side is light than you split that side into two piles of six. If the scale is even than the light one is in your hand. If one side is light than you split that side into two piles of 3 if the scale is even than the one in your hand is the light one. if the scale is light on one side than you take two of the 3 coins left and weigh them. If the scale is even than the coin you have in your hand is the light one. if one side is light than you have your light coin

[Guest]

4 tries, after which I can also tell you whether the coin is heavier or lighter than the other coins...

Here's an interesting variation. Suppose there were two odd coins and that one odd coin weighs half as much as an ordinary coin, and the other odd coin weighs half again more than a regular coin (e.g., regular coins weigh 10, the lighter odd coin weighs 5, and the heavier odd coin weighs 15).

If I had more time I'd write you a shorter description of the solution.

I think the knee jerk reaction is to forget that scales can also balance in addition to showing one side heavier than the other. Thus creating 3 possible outcomes; equal, left heavier, & right heavier. The only obvious way to exploit this is to split the coins into 3 groups. Conversely, since we aren't told heavier/lighter in advance we have to spend a scale measurement to figure that out.

For the first weighing split the 27 coins into 3 piles of 9 coins each; for reference assign each pile a letter: A, B, & C. In this case, I'll let C sit out and place A & B on the scales.

CASE 1 (A = B):

If A = B, that means the odd coin is in C. I don't know if the coin is heavier or lighter yet, I could postpone figuring this out now or I could determine heavier or lighter at this time by another weighing of C against A. Taken to it's logical end, I could get lucky and find the coin in 3 weighings 1/27th of the time but we wouldn't know heavier/lighter then. For sake of consistency with the other two inequality cases, I would do the weighing and note whether C (the one with the odd coin) is heavier or lighter than A.

CASE 2 (A < B):

If A < B, then the odd coin is contained by A or B, but I don't know if it is heavier or lighter yet. I can determine heavier or lighter as well as whether the coin is in A or B with one more weighing of A against C. If A = C, then I know that B contains the odd coin, and that the odd coin is heavier than normal coins. If A < C, then I know that A contains the coin and that it is lighter. Irony has it that A > C isn't really a possibility at this point because it would imply that A has the odd coin and that it is heavier; however by definition C and B weigh identical weights and which contradicts the original weighing result that A < B.

CASE 3 (A > B):

If A > B, I can again weigh A against C and determine which pile has the coin as well as whether the coin is heavier/lighter with similar logic as the last paragraph.

At this point, in two weighings I know two things: 1) whether the coin is heavier/lighter than the rest, and 2) I know which group of 9 coins contains the odd coin.

For the remaining 3rd & 4th weighings I proceed by similarly splitting the group with the coin into 3 groups for the next weighing and leaving one group off the scales. I already know heavier vs lighter, so I'm really just looking for two outcomes: 1) equality, which implies the odd coin is in the third pile, or 2) unbalance, in which case I'm looking for the side with the coin (which I know based on heavier/lighter determined in the second weighing.) Thus the 9 is split out into 3 groups of 3 yielding just one of those groups with the coin. Doing this one more time takes a group of 3 coins and yields the odd coin in 4 weighings, and we know whether it is heavier or lighter.

I find it interesting/amusing that adding anywhere from 1 to 54 more coins would only increase it to 5 weighings.

[Guest]

I got a minimum of 4 and maximum of 7

Minimum

1. Spilt into 3 groups of 9 (let's call the a, b and c respectively, for reference) and weigh A and B. For a minimum number of weighings, these must equal, making the different coin in C

2. spilt C into 3 groups of 3 (let's call them C1, C2 and C3 respectively for reference) and weigh C1 and C2. For a minimum number of weighings, these must equal, making the different coin in C3.

3. C3 will contain 3 coins (let's call them 1,2 and 3 respectively for reference) and weigh 1 and 2. For a minimum number of weighings, these must equal, making 3 the different coin.

4. You then weigh this coin against any other coin and see whether it goes down, so weighs more, or goes up, so weighs less.

Maximum

1. Weigh A and B. Now, if they are unbalanced, then you weigh A with C.

2. If A and C is balanced, the B holds the different coin (dc)If they are unbalanced then A holds the dc. Lets say A=C so B

3. Spilt B - 3 groups of 3 (B1, B2, B3) weigh B1 and B2. Unbalanced, so you weigh B1 and B3.

4. If they are balanced then B2 holds the dc. If they are unbalanced, then B1 holds the dc. Lets say B1 = B3, so B2 holds dc.

5. 3 coins in B2 (1, 2 and 3). Weigh 1 and 2. If they are unbalanced, then you weigh 1 and 3.

6. If they are balanced then 2 holds the dc. If they are unbalanced, then 1 hold the dc. Let's say that 1 = 3, so 2 is the dc.

7. Then you measure 2 with any coin to see whether it ways more or less, so whether it is heavier or lighter.

[Guest]

I got it in 4, a different way, and with some optimism, possibly 2.

27 coins.

Split into two groups. Remainder 1. Weigh the two groups of 13 against eachother. If they balance, the remainder is the coin. (be that the case, remove 1 coin out of one of the groups weighed, and replace it with the remainder coin to determine heavier or lighter)

13 and 13 narrows down to one heavier or lighter group

Split it into two, and take another remainer.

6 and 6 remainder 1

Continue the process

3 and 3 no remainder

1 and 1 remainder 1 <- in this case, you are down to the last 3 coins. If they balance, the remainder is the one. If not, the one that is heavier or lighter, depending on what was determined earlier, is the one.

So the weighing is:

13 v 13

6 v 6

3 v 3

1 v 1 4 times.

[Guest]

You can get it in 3... every time... and its not hard.

First, divide the 27 into 3 piles of 9. Put 9 on one half of the scale and 9 on the other. If one of these groups of 9 weighs less, you will continue with these 9. If they are even, you will continue with the last group of 9 not yet weighed. So at these point we have it narrowed to 9.

Second, divide these 9 into 3 piles of 3. Put 3 on one half of the scale and 3 on the other. If one of these groups of 3 weighs less, you will continue with these 3. If they are even, you will continue with the last group of 3 not yet weighed. Now we have it narrowed down to 3.

Third and last, divide these 3 into 3 piles of 1. Put one on one half of the scale and 1 on the other. If one is lighter you have found your answer. If they are even it must obviously be the last one not weighed.

Solved in 3.

[Prime]

You can get it in 3... every time... and its not hard.

First, divide the 27 into 3 piles of 9. Put 9 on one half of the scale and 9 on the other. If one of these groups of 9 weighs less, you will continue with these 9. If they are even, you will continue with the last group of 9 not yet weighed. So at these point we have it narrowed to 9.

Second, divide these 9 into 3 piles of 3. Put 3 on one half of the scale and 3 on the other. If one of these groups of 3 weighs less, you will continue with these 3. If they are even, you will continue with the last group of 3 not yet weighed. Now we have it narrowed down to 3.

Third and last, divide these 3 into 3 piles of 1. Put one on one half of the scale and 1 on the other. If one is lighter you have found your answer. If they are even it must obviously be the last one not weighed.

Solved in 3.

Per OP, you don't know whether counterfeit weighs more, or less than true coin. So if one group of 9 weighs less than the other, you are left with 18 coins.

[Guest]

Actually, the answer is 1. You make three piles, 13 coins, 13 coins and 1 coin. Weigh the two piles of 13 coins, and if they balance, the odd coin is the heavier/lighter one.

[Guest]

hey... the coin does not have to be light... it can be heavy too... so 3 will not work.... i think 4 tries are required....

1st and 2nd weighs> say first you divide it as a, b, c (9 coins each), and weigh a against b and a against c, the one with an odd coin (less or more weight can be found out) and we will also know if it has more weight or less weight.

3rd weigh> now divide that group into 3 of 3 coins each, say x, y, z. compare x and y. if x=y, z has odd coin. otherwise, according to our knowledge of the nature of the coin (heavy or light), we know which one of x and y contains odd coin.

4th weigh> take that group and weigh one coin against another one. the same argument goes to decide the odd coin (similar to the odd group in the previous step)

[Guest]

the method i mentioned will always work, whatever be the case.

[Guest]

My method would still work in 3 if the coin was heavier... I don't know what you're not getting dude. The answers is 3. Re-read my answer... it works every time, whether the coin is heavier or lighter.

[Prime]

You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g).

The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter?

I have no answer to this question.

Do you?

My method would still work in 3 if the coin was heavier... I don't know what you're not getting dude. The answers is 3. Re-read my answer... it works every time, whether the coin is heavier or lighter.

I have highlighted the relevant parts of the OP (Original Post). Clearly, it implies that you don't know whether counterfeit coin lighter or heavier.

Here is a detailed explanation of how your method could fail:

1. Weigh 9 against 9 and one side is lighter. You took the lighter group of 9 and

2. Weigh 3 vs. 3 with 3 on the side. They weigh the same. So you take 3 on the side and

3. Weigh 1 vs. 1 with 1 on the side. They weigh the same, so you assume (wrongly) that the coin on the side is the counterfeit. Whereas, it was the one the 9 in the heavy group, which you have abandoned.

The answer has been posted throughout this topic many times over. And I have already noted that you can figure out counterfeit out of 39 coins with 4 weighings. But you are not guaranteed to get it out of 27 in less than 4. Three weighings is enough for 12 coins, where you don't know whether counterfeit is heavier, or lighter.

[Guest]

First you put all the coins on the scale your either gonna get 269G or 271G.

Then you keep taking off each coin which would be 10G less each time until you get the one of abnormal weight so I would have to say the answer is once.

[Guest]

Assuming you have a simple scale, my answer is 9.

First divide the coins into 3 piles of 9.

Measure 2 piles at a time until you find the pair that is equal in mass. This can take up to 3 tries, but if you were lucky you could get it on the first try.

Put the piles of coins with equal mass to the side.

Repeat the process this time with 3 piles of 3 coins, maximum 3 measurements.

Repeat the process this time with the 3 remaining coins, find the two that are equal, the remaining coin has a different mass.

If you are a very luck person you coulp potentially only have to make 4 meaurements, 3 to rule out equal pairs and 1 to compare the final coin to the rest.

9 is the most you would have to do the efficient way, and at any point you can observe is the unequal pile has more or less mass.

Six is the minimum for exact weight of finding the coin

1. Divide 27 into 3 piles of 9 coins each.

Weigh two piles against each other.

If weight is same, the third pile has the coin of different weight.

2. If weight is different, take one pile off the weighing machine and put the other pile of 9 coins.

Again weigh against each other.

If weights are same, then the pile you took off has the coin. If not, then the pile on the machine has the coin

So in 1 or 2 weighings you got the pile of 9 coins.

3. Similarly, divide the 9 coins into 3 piles of 3 coins each

Weigh two piles against each other.

If weight is same, the third pile has the coin of different weight.

4. If weight is different, take one pile off the weighing machine and put the other pile of 3 coins.

Again weigh against each other.

If weights are same, then the pile you took off has the coin. If not, then the pile on the machine has the coin

So in 3 or 4 weighings you got the pile of 3 coins.

5. Similarly, divide the 3 coins into 3 piles of 1 coin each

Weigh two piles against each other.

If weight is same, the third pile has the coin of different weight.

6. If weight is different, take one pile off the weighing machine and put the other pile of 1 coins.

Again weigh against each other.

If weights are same, then the pile you took off has the coin. If not, then the pile on the machine has the coin

So in 5 or 6 weighings you got the coin with different weight.

[Guest]

ok guys...heres my point of view on this one... though this sounds difficult to perform...

This will tell you in one chance...Asssume you have a Tall Transparent column of water... now throw all the coins (very important) all at the same time into the water... one of those coins will fall fraction later or before all the rest hit the bottom... thats the one baby...

[Guest]

It only takes two weighing s!!!

Here is how- First divide the coins into the three piles of nine which I will call A,B, and C. Then weigh the first two piles, piles A and B. The scale will either be 1.-even or 2.-uneven

1. If the scale is even then you take out pile B and put pile C in. If the scale on pile c's side goes down then the coin weighs 11g and if it goes up it only weighs 9g.

2. If the scale is uneven then take out pile B and put pile in pile C. If the scale went down for pile B and it goes down for pile C then we know that pile A contains the coin weighing 9g. If the scale goes up for pile B and C then we know pile A contains a coin weighing 11g. If you put pile C in and the scale becomes even then you know that pile B has a coin weighing either 9 or 11 g depending on which way it tipped on the initial weighing.

[Guest]

I came up with what, to me, seemed like an even simpler explanation.

You have 27 coins...

Divide them into two groups of 13, holding one coin out.

the measurements on the scale will read 130g for one group and either 130, 131, or 129 for the second. Disregard the group with the weight of 130g.

Assuming you didn't pull the odd coin, again split the coins in the lighter/heavier group in half once again leaving out one coin

This time the scales would read 60g for one group and 60/61/59 for the other

The third time, you'll have no need to remove a coin, so just weigh the two groups of three.

Same as before the scales will read 30g for one and either 30/31/29 for the third.

You're now down to three coins. Hold one out and weigh the final two, and voila!

Edited September 23, 2009 by Shachi