80 identical coins

[Guest]

You have 80 identical coins which are all authentic, and also one counterfeit which seems identical to the others but is slightly heavier. You have three balance scales, but one of them is broken and gives random answers. How do you find the counterfeit coin in 7 weighings?

[CaptainEd]

I assume that the broken balance can give accurate readings, and can give any answer, can say heavy is heavy, even, or light, and the same for equal.

If two scales give the same readings with the same coins, the reading is accurate, but one of them could still be the broken scale.

If two scales give different readings with the same coins, the third scale is accurate. We can use it for the rest of the weighings.

call the scales S1, S2, S3, and the weighings W1, W2...W7

W1: weigh 27 against 27 in Scale S1, leave 27 aside

W2: weigh the same 27 and 27 in S2.

If they agree, take the low 27 or the third 27:

-- W3: weigh 9 against 9 in S1, leave 9 aside

-- W4: weigh same 9 vs 9 in S2

-- If they agree, take the low 9 or the third 9:

------W5: weigh 3 vs 3 in S1

------W6: weigh same 3 vs 3 in S2

------If they agree, take the low 3 or the third 3:

---------W7: weigh 1 vs 1 in S1

---------W8: weigh 1 vs 1 in S2

---------If they agree, take the low 1 or the third 1

---------If they disagree,

------------W9: weigh same 1 vs 1 in S3, and believe the result (take low 1 or third 1)

------If they disagree,

---------W7: weigh same 3 vs 3 in S3, take low 3 or third 3

---------W8: weigh 1 vs 1 in S3, believe the result

---If they disagree,

------W5: weigh same 9 vs 9 in S3, take the low 9 or the third 9:

------W6: weigh 3 vs 3 in S3, take the low 3 or third 3:

------W7: weigh 1 vs 1 in S3, believe the result

If they disagree,

-- W3: weigh same 27 vs 27 in S3, take the low 27 or the third 27:

-- W4: weigh 9 vs 9 in S3, take the low 9 or the third 9

-- W5: weigh 3 vs 3 in S3, take the low 3 or the third 3

-- W6: weigh 1 vs 1 in S3, believe the result

[Guest]

I assume that the broken balance can give accurate readings, and can give any answer, can say heavy is heavy, even, or light, and the same for equal.

If two scales give the same readings with the same coins, the reading is accurate, but one of them could still be the broken scale.

If two scales give different readings with the same coins, the third scale is accurate. We can use it for the rest of the weighings.

call the scales S1, S2, S3, and the weighings W1, W2...W7

W1: weigh 27 against 27 in Scale S1, leave 27 aside

W2: weigh the same 27 and 27 in S2.

If they agree, take the low 27 or the third 27:

-- W3: weigh 9 against 9 in S1, leave 9 aside

-- W4: weigh same 9 vs 9 in S2

-- If they agree, take the low 9 or the third 9:

------W5: weigh 3 vs 3 in S1

------W6: weigh same 3 vs 3 in S2

------If they agree, take the low 3 or the third 3:

---------W7: weigh 1 vs 1 in S1

---------W8: weigh 1 vs 1 in S2

---------If they agree, take the low 1 or the third 1

---------If they disagree,

------------W9: weigh same 1 vs 1 in S3, and believe the result (take low 1 or third 1)

------If they disagree,

---------W7: weigh same 3 vs 3 in S3, take low 3 or third 3

---------W8: weigh 1 vs 1 in S3, believe the result

---If they disagree,

------W5: weigh same 9 vs 9 in S3, take the low 9 or the third 9:

------W6: weigh 3 vs 3 in S3, take the low 3 or third 3:

------W7: weigh 1 vs 1 in S3, believe the result

If they disagree,

-- W3: weigh same 27 vs 27 in S3, take the low 27 or the third 27:

-- W4: weigh 9 vs 9 in S3, take the low 9 or the third 9

-- W5: weigh 3 vs 3 in S3, take the low 3 or the third 3

-- W6: weigh 1 vs 1 in S3, believe the result

That was my first approach as well. It can be done in only 7 weighings though

[Guest]

weigh 40 against 40 in scale 1.

if neither side heavy, you have the random scale.

---weigh 27 against 27 in scale 2.

---weigh 9 against 9 based on heavy.

---weigh 3 against 3

---weigh 1 against 1.

if one side heavy weigh 40 against 40 in scale 2. if they agree there are two possibilities. either 1 is random and recorded accurately, or both are accurate. a quick way to verify, no matter the outcome, would be to weigh 40 against 40 in the last scale. then follow the above, based on results.

i suspect there is a way to solve this in fewer moves.

Edited September 19, 2011 by phillip1882

[Guest]

hmm the above doesn't work in all cases. if all three agree, we still don't know which scale is random.

to be honest, i dont see a way to garentee a solution now.

you only have 3 coins, one is heavy but you don't know which one. you have 3 scales, and one answers randomly.

all three scales look alike obviously.

how would you detect the bad scale?

of course, probabilistically, if you wiegh enough times, you can detect it. simply wiegh two coins in each scale untill you get a different answer. but it could take you years! :-)

[Guest]

weigh 27 agianst 27 in two scales.

based on results, wiegh 9 agianst 9.

then 3 against 3, and 1 against 1.

in each case, it takes two weighings.

you can make it quite probable of only taking 7 by mixing up which two scales you use, but there will be some case requiring 8 as far as i can see.

[plasmid]

If I have nine coins (one of which is heavy) with three scales (one of which gives random answers), then I have an algorithm that will either identify the heavy coin AND the broken scale in four weighings, or will only identify the heavy coin in three weighings.

To apply this to 81 coins, group coins 1-9, 10-18, 19-27, etc into groups and treat each group as if it were a single coin. After applying this algorithm, you will either identify the group with a heavy coin AND the broken scale in four weighings (in which case it's easy to figure out the heavy coin within that group in three weighings), or you will have identified the group with a heavy coin in three weighings without knowing which scale is broken (in which case you can just apply the algorithm again to the nine coins of that group).

It is much easier for me to show this in table form than explain it entirely in words. The two left hand columns list all possible combinations of which coin is heaviest and which balance is broken. Columns after that give the outcome of each weighing, either L if the left pan is heavier, R if the right pan is heavier, E if the pans are even, or ? if the scale is broken.

The first weighing will be of coins 1, 2, 3 on the left pan against 4, 5, 6 on the right pan done on scale #1 (denoted "123 vs 456 on 1" on the worksheet). If the left pan is heaviest, then go to the next column (If L; 145 vs 267 on 2) meaning you'll weigh coins 1, 4, 5 against 2, 6, 7 on scale #2, and you can see all possible outcomes for the second weighing covering the cases where the first weighing could have had the left pan be heavier. If the first weighing had the left pan heavier and the second weighing had the left pan heavier then look for the column "If LL" to describe the next weighing, or if the first pan had the left pan heavier and the second weighing had the pans come out even then look for the column "If LE" to describe the next weighing.

Of course, you might have noticed that there is no "If LE" column...

This worksheet doesn't exhaustively cover every case. That is because I am lazy. And because there is a lot of symmetry between the cases. For example, if the first weighing came out even, then you could simply rub off the numbers on coins 7, 8, 9 and replace them with numbers 1, 2, 3 (and vice-versa to relabel the original coins 1, 2, 3 to now be 7, 8, 9) and proceed as if the left pan were heavier. Similarly, if the second weighing came out even, you could swap the labels on coin 1 and coin 3, and switch the labels on coins 4, 5 and coins 8, 9, and proceed as if the left pan were heavier. Since this is brainden, I'll leave it to you to see that all possible cases can be covered by symmetry.

______________First_weigh______If_L_____________If_LL__________If_LLL

Heavy_Broken__123_vs_456_on_1__145_vs_267_on_2__23_vs_45_on_3__2_vs_3_on_3

1_____1_______?________________L________________E

1_____2_______L________________?________________E

1_____3_______L________________L________________E

2_____1_______?________________R

2_____2_______L________________?________________L______________L

2_____3_______L________________R

3_____1_______?________________E

3_____2_______L________________?________________L______________R

3_____3_______L________________E

4_____1_______?________________L________________R

4_____2_______R

4_____3_______R

5_____1_______?________________L________________R

5_____2_______R

5_____3_______R

6_____1_______?________________R

6_____2_______R

6_____3_______R

7_____1_______?________________R

7_____2_______E

7_____3_______E

8_____1_______?________________E

8_____2_______E

8_____3_______E

9_____1_______?________________E

9_____2_______E

9_____3_______E

(Apologies for having to put all those underscores in the worksheet. Otherwise it wouldn't show up aligned correctly, even if I put everything in a code box.)

[Guest]

I wrote this quickly, hope it makes sense:

Divide into 4 groups, G1, G2, G3, G4 such that groups G1,G2, & G3 have equal number of coins,

W= weigh number, S=scale number B=balance H=heavy L=Light U=unbalanced

W1 - S1 - G1 vs G2

W2 - S2 - G1 vs G3

W3 - S3 - G2 vs G3

Options are:

1. All 3 B, bad coin must be in G4

2. 2 B, 1 U. then bad scale has given bad reading but do not know which one.And, the group common to the 2 balanced scales has been identified as all good coins. Let's assume S1 and S2 balanced and S3 shows G2 H, then G1 and G3 have all good coins. and options are S3 is bad and conterfeit is in G4 or S3 is good and Conterfeit is in G2

3, 1 B, 2 U. then coins in group G4 are good coins. And, if the U in both cases indicate the same group H then the conterfeit is in the H group. If different groups are H, one of these has the Conterfeit and the. one of these two scales are broken and G4 & two groups on the B scale are good coins if the U is different, then one of those two scales must be the bad 1 so use scale that balanced for further wieghings and the bad coin must be in the common group.

To summarize results to this point for what each option yields:

1. Bad coin in G4 but do not know which scale is bad

2. Bad coin is either in Heavey group or G4, if it is in G4 then U scale is bad. If it is in H group from U scale, then the bad scale is the other one that weighed the H group.

3a. same H group so H group has bad coin but no knowledge of bad scale.

3b. Different group shows H, then balanced scale is good scale and group not on balanced scale has bad coin.

So after three weighings we can determine which group of the 4 groups has the counterfeit, No time to continue now, propose even groups of 20 and G4 having 21. Would like to know if this is right approach.

[bhramarraj]

There are 81 coins, of which one is heavier coin.

Make three groups A,B,C each of 27 coins. Weigh A & B in first balance. The result obtained in first weighing, if correct, will decide the heavier group.

Divide this supposedly heavier group and also one lighter group, each in two groups having 14 & 13 coins each. For convenience name them (14H, 13H) & (14L, 13L) respectively. Weigh them in second balance.

Then second weighing, in second balance, may give following results:

[1] 14H + 13L = 13H + 14L

[2] 14H + 13L > or < 13H +14L

[1] If we get result [1], then either first or the second balance is incorrect, which decides THIRD BALANCE AS CORRECT.

Then third weighing in third correct balance will decide heavier group of 27 coins, fourth weighing will decide heavier group of 9 coins,fifth weighing will decide heavier group of 3 coins, and finally sixth weighing will decide the one heavier coin.

[2] If we get result [2], then both results are correct. So we do not know the correct balance, but we know out of 14H or 13H, which one is heavier.

Now divide these 14 or 13 coins in three groups of 5, 5, and 4 or three coins.

Weigh two groups of 5 coins in first balance. Assuming this third weighing in first balance as correct, will decide the probable group containing heavier coin. divide this heavier group in two groups containing (3H, 2H), or (2H, 2H) or (2H, 1H); and divide one lighter group as (3L,2L), or (2L, 2L) or (2L, 1L).

Fourth weighing of (3H + 2L) & (3L + 2H) or (2H + 2L) & (2H + 2L) or (2H + 1L) & (2L + 1H), in second balance, will decide the following;

[1] If balanced, then one of the balance out of first and second is incorrect. Then fifth weighing in third correct scale will decide the heavier group 5 or 4 or 3 coins. Sixth & seventh weighing will decide one heavier coin.

[2] If unbalanced, then we know the heavier group of 3 or 2 or 1 coin, but status of the balances is not known.

Now we have max 3 coins. Fifth and sixth weighing will decide the one heavier coin.

Edited September 26, 2011 by bhramarraj

[bhramarraj]

There are 81 coins, of which one is heavier coin.

Make three groups A,B,C each of 27 coins. Weigh A & B in first balance. The result obtained in first weighing, if correct, will decide the heavier group.

Divide this supposedly heavier group and also one lighter group, each in two groups having 14 & 13 coins each. For convenience name them (14H, 13H) & (14L, 13L) respectively. Weigh them in second balance.

Then second weighing, in second balance, may give following results:

[1] 14H + 13L = 13H + 14L

[2] 14H + 13L > or < 13H +14L

[1] If we get result [1], then either first or the second balance is incorrect, which decides THIRD BALANCE AS CORRECT.

Then third weighing in third correct balance will decide heavier group of 27 coins, fourth weighing will decide heavier group of 9 coins,fifth weighing will decide heavier group of 3 coins, and finally sixth weighing will decide the one heavier coin.

[2] If we get result [2], then both results are correct. So we do not know the correct balance, but we know out of 14H or 13H, which one is heavier.

Now divide these 14 or 13 coins in three groups of 5, 5, and 4 or three coins.

Weigh two groups of 5 coins in first balance. Assuming this third weighing in first balance as correct, will decide the probable group containing heavier coin. divide this heavier group in two groups containing (3H, 2H), or (2H, 2H) or (2H, 1H); and divide one lighter group as (3L,2L), or (2L, 2L) or (2L, 1L).

Fourth weighing of (3H + 2L) & (3L + 2H) or (2H + 2L) & (2H + 2L) or (2H + 1L) & (2L + 1H), in second balance, will decide the following;

[1] If balanced, then one of the balance out of first and second is incorrect. Then fifth weighing in third correct scale will decide the heavier group 5 or 4 or 3 coins. Sixth & seventh weighing will decide one heavier coin.

[2] If unbalanced, then we know the heavier group of 3 or 2 or 1 coin, but status of the balances is not known.

Now we have max 3 coins. Fifth and sixth weighing will decide the one heavier coin.

Hi David... ! is there any other simpler solution.......?

Edited September 27, 2011 by bhramarraj

[plasmid]

There are 81 coins, of which one is heavier coin.

Make three groups A,B,C each of 27 coins. Weigh A & B in first balance. The result obtained in first weighing, if correct, will decide the heavier group.

Divide this supposedly heavier group and also one lighter group, each in two groups having 14 & 13 coins each. For convenience name them (14H, 13H) & (14L, 13L) respectively. Weigh them in second balance.

Then second weighing, in second balance, may give following results:

[1] 14H + 13L = 13H + 14L

[2] 14H + 13L > or < 13H +14L

[1] If we get result [1], then either first or the second balance is incorrect, which decides THIRD BALANCE AS CORRECT.

Then third weighing in third correct balance will decide heavier group of 27 coins, fourth weighing will decide heavier group of 9 coins,fifth weighing will decide heavier group of 3 coins, and finally sixth weighing will decide the one heavier coin.

[2] If we get result [2], then both results are correct. So we do not know the correct balance, but we know out of 14H or 13H, which one is heavier.

I think the statement right above this is not always necessarily true. If scale 1 is accurate, and scale 2 is broken, you would not know for sure that the group of 14 or 13 that the second scale identified as heavier really is the heavier group.

Now divide these 14 or 13 coins in three groups of 5, 5, and 4 or three coins.

Weigh two groups of 5 coins in first balance. Assuming this third weighing in first balance as correct, will decide the probable group containing heavier coin. divide this heavier group in two groups containing (3H, 2H), or (2H, 2H) or (2H, 1H); and divide one lighter group as (3L,2L), or (2L, 2L) or (2L, 1L).

Fourth weighing of (3H + 2L) & (3L + 2H) or (2H + 2L) & (2H + 2L) or (2H + 1L) & (2L + 1H), in second balance, will decide the following;

[1] If balanced, then one of the balance out of first and second is incorrect. Then fifth weighing in third correct scale will decide the heavier group 5 or 4 or 3 coins. Sixth & seventh weighing will decide one heavier coin.

[2] If unbalanced, then we know the heavier group of 3 or 2 or 1 coin, but status of the balances is not known.

Now we have max 3 coins. Fifth and sixth weighing will decide the one heavier coin.

Comment is added in blue.

[plainglazed]

1. Using scale 1, compare 27 vs 27 coins.

2. Using scale 2, compare 12 vs 12 of the 27 coins indicated as heavy from the first weighing.

If = either scale 1 or scale 2 can still be either random or read true.

3. Using scale 2 again, compare the 3 not yet weighed from the 27 indicated as heavy from the first weighing vs three from the 24 from the second weighing.

If = either one of scales 1 or 2 is random and wrong so scale 3 must read true. Four more weighings using scale 3 reading true (27 vs 27, 9 vs 9, 3 vs 3, 1 vs 1) can determine the heavy coin.

If > or < then the first weighing must be accurate (tho scale 1 is either random or reads true) as at most only one of two scales can be wrong.

If < scale 2 is random and wrong as it contradicted itself so scales 1 and 3 must read true. Four more weighings using scale 1 or 3 reading true (27 vs 27, 9 vs 9, 3 vs 3, 1 vs 1) can determine the heavy coin.

4. Using scale 3, repeat the third weighing to confirm

If weighing 4 does not read the same as weighing 3, one of scales 2 or 3 is random and wrong. Three more weighings using scale 1 reading true (9 vs 9, 3 vs 3, 1 vs 1) can determine the heavy coin.

If weighing 4 and 3 are the same, the heavy coin is confirmed in the three indicated as heavy.

5. 6. 7. Using different scales compare 1 vs 1 of the three indicated as heavy.

Whichever reading occurs two or three times must be accurate and determines the heavy coin.

2. Using scale 2, compare 12 vs 12 of the 27 coins indicated as heavy from the first weighing.

If > or < the first weighing must be accurate (tho scale 1 is either random or reads true) as at most only one of the two scales can be wrong.

3. Using scale 3 compare 6 vs 6 of the 12 coins indicated as heavy from the second weighing.

If = then one of scales 2 or 3 is random and wrong so scale 1 must read true. Three more weighings using scale 1 reading true (9 vs 9, 3 vs 3, 1 vs 1) can determine the heavy coin.

If > or < the second weighing must be accurate as at most only one of the two scales can be wrong.

4. Using scale 1, compare 3 vs 3 of the 6 coins indicated as heavy from the third weighing.

If = then one of scales 3 or 1 is random and wrong so scale 2 must read true. Three more weighings using scale 2 reading true (6 vs 6, 3 vs 3, 1 vs 1) can determine the heavy coin.

If > or < the third weighing must be accurate as at most only one of the two scales can be wrong.

5. Using scale 2, compare 2 vs 1,X (known normal coin) of the 3 coins indicated as heavy from the fourth weighing.

If = then one of scales 2 or 1 is random and wrong so scale 3 must read true. Two more weighings using scale 3 reading true (3 vs 3, 1 vs 1) can determine the heavy coin.

If > or < the fourth weighing must be accurate as at most only one of the two scales can be wrong.

Label the three coins now known to contain the heavy coin a,b,c

5. Using scale 2 compare a,b vs c,X (already indicated as > or <)

6. Using scale 3 compare b,c vs a,X

7. Using scale 1 compare c,a vs b,X

If a is heavy, true readings of 5. 6. 7. would be >,<,>

If b is heavy, true readings of 5. 6. 7. would be >,>,<

If c is heavy, true readings of 5. 6. 7. would be <,>,>

Quick inspection will show that if any one of these readings is changed from a random scale reading incorrectly, the remaining two readings are still unique thus the heavy coin can be determined.

EDITS in red

Hope I didn't miss something. Easy to get confused with the random scale. A most excellent puzzle shakingdavid! Five stars from me.

[bhramarraj]

Thanks Plasmid, you are absolutely right. I overlooked this.

So I am again posting a modified solution, and awaiting valuable comments.

Divide 81 coins in three groups 27A, 27B, and 27C.

W1___S1: Compare 27A and 27B in S1 balance. One of the three following results is possible:

[1] 27A = 27B, [2] 27A > 27B, [3] 27A < 27B

:

Divide 27A and 27C, each into two groups 14A & 13A, and 14C & 13C.

W2___S2: Compare 14A + 13C and 13A + 14C in S2 balance, to get one of the three following results:

[1] 14A + 13C = 13A + 14C, [2] 14A + 13C > 13A + 14C,

[3] 14A + 13C < 13A + 14C

S1- [1] & S2 - [1]:

One of the balances, S1 or S2, is definitely broken. So S3 must be definitely genuine one. Hence we find the heavier coin by weighing in S3 as under:

W3__S3: find heavier group of 27 coins,

W4__S3: find heavier group of 9 coins,

W5__S3: find heavier group of 3 coins,

W6__S3: find heavier one coin.

S1- [1] & S2 - [2]:

Both weighing may be correct. Broken balance is also not known.

Divide 27B into two groups 14B & 13B.

W3__S3: Compare 14B + 13C and 13B + 14C in S2 balance, to get one of the three following results:

[1] 14B + 13C = 13B + 14C, [2] 14B + 13C > 13B + 14C,

[3] 14B + 13C < 13B + 14C

S1- [1], S2 - [2] & S3 - [1]:

One of the balances, S1 & S3, is definitely broken. So S2 must be definitely genuine one. Hence we find the heavier coin by weighing in S2 as following:

W4__S2: find heavier group of 27 coins,

W5__S2: find heavier group of 9 coins,

W6__S2: find heavier group of 3 coins,

W7__S2: find heavier one coin.

S1- [1], S2 - [2] & S3 - [2]:

One of the balances, S2 & S3, is definitely broken. So S1 must be definitely genuine one. W1__S1 already indicated that 27C contain heavier coin. Hence we find the heavier coin by weighing in S1 as following:

W4__S1: find heavier group of 9 coins,

W5__S1: find heavier group of 3 coins,

W6__S1: find heavier one coin.

S1- [1], S2 - [2] & S3 - [3]:

One balance out of S2 & S3 is broken. So S1 is OK. further weighing in S1 will reveal heavier coin as above.

Same way with other possible combination of results, we are able to find the heavier one coin in maximum 7 weighing.

[bhramarraj]

Thanks Plasmid, you are absolutely right. I overlooked this.

So I am again posting a modified solution, and awaiting valuable comments.

Divide 81 coins in three groups 27A, 27B, and 27C.

W1___S1: Compare 27A and 27B in S1 balance. One of the three following results is possible:

[1] 27A = 27B, [2] 27A > 27B, [3] 27A < 27B

:

Divide 27A and 27C, each into two groups 14A & 13A, and 14C & 13C.

W2___S2: Compare 14A + 13C and 13A + 14C in S2 balance, to get one of the three following results:

[1] 14A + 13C = 13A + 14C, [2] 14A + 13C > 13A + 14C,

[3] 14A + 13C < 13A + 14C

S1- [1] & S2 - [1]:

One of the balances, S1 or S2, is definitely broken. So S3 must be definitely genuine one. Hence we find the heavier coin by weighing in S3 as under:

W3__S3: find heavier group of 27 coins,

W4__S3: find heavier group of 9 coins,

W5__S3: find heavier group of 3 coins,

W6__S3: find heavier one coin.

S1- [1] & S2 - [2]:

Both weighing may be correct. Broken balance is also not known.

Divide 27B into two groups 14B & 13B.

W3__S3: Compare 14B + 13C and 13B + 14C in S3 balance, to get one of the three following results:

[1] 14B + 13C = 13B + 14C, [2] 14B + 13C > 13B + 14C,

[3] 14B + 13C < 13B + 14C

S1- [1], S2 - [2] & S3 - [1]:

One of the balances, S1 & S3, is definitely broken. So S2 must be definitely genuine one. Hence we find the heavier coin by weighing in S2 as following:

W4__S2: find heavier group of 27 coins,

W5__S2: find heavier group of 9 coins,

W6__S2: find heavier group of 3 coins,

W7__S2: find heavier one coin.

S1- [1], S2 - [2] & S3 - [2]:

One of the balances, S2 & S3, is definitely broken. So S1 must be definitely genuine one. W1__S1 already indicated that 27C contain heavier coin. Hence we find the heavier coin by weighing in S1 as following:

W4__S1: find heavier group of 9 coins,

W5__S1: find heavier group of 3 coins,

W6__S1: find heavier one coin.

S1- [1], S2 - [2] & S3 - [3]:

One balance out of S2 & S3 is broken. So S1 is OK. further weighing in S1 will reveal heavier coin as above.

Same way with other possible combination of results, we are able to find the heavier one coin in maximum 7 weighing.

There is a typo error in the 18th line from top 'W3__S3: Compare 14B + 13C and 13B + 14C in S2 balance.........' this may please be read as 'W3__S3: Compare 14B + 13C and 13B + 14C in S3 balance........'.

I regret for the inconvenience caused.

I think this is the correct simplest solution.

[plainglazed]

Thanks Plasmid, you are absolutely right. I overlooked this.

So I am again posting a modified solution, and awaiting valuable comments.

Divide 81 coins in three groups 27A, 27B, and 27C.

W1___S1: Compare 27A and 27B in S1 balance. One of the three following results is possible:

[1] 27A = 27B, [2] 27A > 27B, [3] 27A < 27B

:

Divide 27A and 27C, each into two groups 14A & 13A, and 14C & 13C.

W2___S2: Compare 14A + 13C and 13A + 14C in S2 balance, to get one of the three following results:

[1] 14A + 13C = 13A + 14C, [2] 14A + 13C > 13A + 14C,

[3] 14A + 13C < 13A + 14C

S1- [1] & S2 - [1]:

One of the balances, S1 or S2, is definitely broken. So S3 must be definitely genuine one. Hence we find the heavier coin by weighing in S3 as under:

W3__S3: find heavier group of 27 coins,

W4__S3: find heavier group of 9 coins,

W5__S3: find heavier group of 3 coins,

W6__S3: find heavier one coin.

S1- [1] & S2 - [2]:

Both weighing may be correct. Broken balance is also not known.

Divide 27B into two groups 14B & 13B.

W3__S3: Compare 14B + 13C and 13B + 14C in S2 balance, to get one of the three following results:

[1] 14B + 13C = 13B + 14C, [2] 14B + 13C > 13B + 14C,

[3] 14B + 13C < 13B + 14C

S1- [1], S2 - [2] & S3 - [1]:

One of the balances, S1 & S3, is definitely broken. So S2 must be definitely genuine one. Hence we find the heavier coin by weighing in S2 as following:

W4__S2: find heavier group of 27 coins,

W5__S2: find heavier group of 9 coins,

W6__S2: find heavier group of 3 coins,

W7__S2: find heavier one coin.

S1- [1], S2 - [2] & S3 - [2]:

One of the balances, S2 & S3, is definitely broken. So S1 must be definitely genuine one. W1__S1 already indicated that 27C contain heavier coin. Hence we find the heavier coin by weighing in S1 as following:

W4__S1: find heavier group of 9 coins,

W5__S1: find heavier group of 3 coins,

W6__S1: find heavier one coin.

S1- [1], S2 - [2] & S3 - [3]:

One balance out of S2 & S3 is broken. So S1 is OK. further weighing in S1 will reveal heavier coin as above.

Same way with other possible combination of results, we are able to find the heavier one coin in maximum 7 weighing.

W1___S1 - [1] 27A = 27B

W2___S2 - [2] 14A + 13C > 13A + 14C

W3___S3 - [2] 14B + 13C > 13B + 14C

What if the heavy coin is in the group of 13C?

[bhramarraj]

W1___S1 - [1] 27A = 27B

W2___S2 - [2] 14A + 13C > 13A + 14C

W3___S3 - [2] 14B + 13C > 13B + 14C

What if the heavy coin is in the group of 13C?

W1___S1 - [1] 27A = 27B

W2___S2 - [2] 14A + 13C > 13A + 14C

W3___S3 - [2] 14B + 13C > 13B + 14C

What if the heavy coin is in the group of 13C?

I again overlooked.

Rethinking is needed,.

There must be some simpler solution to this puzzle.I will again try....

[Guest]

Divide group into 3 sets A, B & C (making 27 coins per group) W1, W2 …W7 (Weighing Numbers)

W1. W2. W3. Weigh set A of 27 coins in all balances (Makes 3 weighing already). With this you eliminate the damage balance. Since the elimination of inaccurate balance, I can say that all next weighing are accurate.

W4. Weigh set B or C of 27 coins in the any of the remaining balance. If set B is heavier than A, we can say that set C contains authentic coins. If set B is equal to set A then we can say that set C is the heavier set of coins. And if set B is less than set A, then set A has the counterfeit coin with it.

Now we only got 27 coins remaining. Again divide the group into 3 (making 9 coins per group), we call them set D, E & F

W5. Put set D & E in the two remaining balance. Compare. If D & E is equal then F is heavier, if they have different mass, whichever is higher has the counterfeit coin.

Now we only got 9 coins remaining. Again divide the group into 3 (making 3 coins per group), we call them set G, H & J

W6. Put set G & H in the two remaining balance. Compare. If G & H is equal then J is heavier, if they have different mass, whichever is higher has the counterfeit coin.

Now we only got 3 coins remaining. Again divide the group into 3 (making 1 coin per group), we call them set K, L & M

W7. Put set K & L in the two remaining balance. Compare. If K & L is equal then M is heavier, if they have different mass, whichever is higher is the counterfeit coin.

Yes Tricky but true 7 weighing.

Edited October 4, 2011 by beryong27

[CaptainEd]

Beryong, I think your first step is optimistic--you believe that with 3 weighings, you can find the inaccurate balance. The OP merely states that it gives "random" results, but it does NOT guarantee that it always gives wrong results. What if Scale 2 is the bad one, and the first time you use it, it gives correct results? Then, at the end of the first 3 weighings, you find that you have totally consistent information--you know which 27 are heavy, but you don't know which scale is bad. Now how do you finish in only 4 weighings?

[Guest]

Beryong, I think your first step is optimistic--you believe that with 3 weighings, you can find the inaccurate balance. The OP merely states that it gives "random" results, but it does NOT guarantee that it always gives wrong results. What if Scale 2 is the bad one, and the first time you use it, it gives correct results? Then, at the end of the first 3 weighings, you find that you have totally consistent information--you know which 27 are heavy, but you don't know which scale is bad. Now how do you finish in only 4 weighings?

I've preferred to be optimistic because if this inaccurate balance gives you random reading then theres no way to solve this puzzle, what if this balance gives you 10 consecutive correct reading before giving you a wrong reading, there's no way to solve this rather than to be OPTIMISTIC. Maybe the OP should reveal and explain it.

[plasmid]

Um, I already solved it in post 7?

[Guest]

Hello again

plasmid has already solved the puzzle. The inaccurate balance gives random readings, so we can not assume that the readings are wrong. But the puzzle can still be solved, because you don't need to identify which balance is broken, only which coin is a forgery.

[bhramarraj]

I found the wrong step. Please cmment on my third modified solution in spoiler.

Well, after reviewing my earlier posts I came to know where I was wrong footed. Here is the modified solution.

Divide 81 coins in three groups 27A, 27B, & 27C.

W1S1:

Compare 27A & 27B in S1 balance. Three conditions are possible:

(a1) 27A = 27B; (b1) 27A > 27B; & (c1) 27A < 27B.

If we get (a1), then 27C would have heavier coin. So in next weighing in balance S2, if we get result which doesn’t show 27C side heavier, then it will be confirmed that one of the balance between S1 & S2 is broken i.e. S3 is OK. Then there will be no problem in finding the heavier coin within seven weighing.

So we will consider only the possibilities of getting those results in W2S2 which show 27C containing heavier coin.

W2S2:

Divide 27B & 27C each in two groups - 14B, 13B & 14C & 13C. Compare them in S2. Three conditions are possible:

(b2) 14B + 13C > 13B + 14C; (c2) 14B + 13C < 13B + 14C.

If we get result (b2), then 13C would have heavier coin; so in next weighing W3S3, if we get result which does not show 13C side heavier, then it will be confirmed that one of the balance between S2 & S3 is broken i.e. S1 is OK. Then there will be no problem in finding the heavier coin within seven weighing.

So we will consider only the possibilities of getting those results in W3S3 which show 13C containing heavier coin.

W3S3:

Divide 13C in groups 7C, 6C and 13B in 7B, 6B. Comparing them in S3 may give one of the following results:

(b3) 7C + 6B > 7B + 6C; (c3) 7C + 6B < 7B + 6C;

If we get result (b3), then 7C would have heavier coin; so in next weighing W4S1, if we get result which does not show 7C side heavier, then it will be confirmed that one of the balance between S1 & S3 is broken i.e. S2 is OK. Then there will be no problem in finding the heavier coin within seven weighing.

So we will consider only the possibilities of getting those results in W4S1 which show 7C containing heavier coin.

W4S1:

Divide 7C in groups 4C, 3C and 13B in 4B, 3B. Comparing them in S1 may give one of the following results:

(b4) 4C + 3B > 4B + 3C; (c4) 4C + 3B < 4B + 3C;

If we get result (b4), then 4C would have heavier coin; so in next weighing W5S2, if we get result which does not show 4C side heavier, then it will be confirmed that one of the balance between S1 & S3 is broken i.e. S2 is OK. Then there will be no problem in finding the heavier coin within seven weighing.

So we will consider only the possibilities of getting those results in W5S2 which show 4C containing heavier coin.

W5S2:

Divide 4C in groups 2C1, 2C2 and 4B in 2B1, 2B2. Comparing them in S2 may give one of the following results:

(b5) 2C1 + 2B1 > 2B2 + 2C2; (c5) 2C1 + 2B1 < 2B2 + 2C2;

If we get result (b5), then 2C1 would have heavier coin; so in next weighing W6S3, if we get result which does not show 2C1 side heavier, then it will be confirmed that one of the balance between S2 & S3 is broken i.e. S1 is OK. Then there will be no problem in finding the heavier coin within seven weighing.

So we will consider only the possibilities of getting those results in W6S3 which show 2C1 containing heavier coin.

W6S3:

Compare two coins of 2C1 in S3 balance, one of the coin will show heavier. Heavier coin may be confirmed in W7S1.

Same procedure will follow for other results (b1), (c1) also only difference will be that we will have to take 27A or 27B whichever heavier, in place of 27C.

[plainglazed]

If W5S2 reads = then either W5S2 or W4S1 must be random and wrong so consequently W3S3 must be accurate and read true: the heavy coin in 7C. two more weighings using S3 with this knowledge is not enough to determine the heavy coin.

If W6S3 and W7S1 are different, how do you know which is a true reading and which is random and wrong?

am impressed with your perserverance and I think you have the right concept. used similar myself with a little trick to have more room in the end for final determination.

EDIT: corrected poor/hasty reasoning

[Guest]

A Most Unique Solution.

I'm not sure if this actually counts (even though it works).

1. Put scales one and two on each side of scale three.

2. Put 20 coins in each side of scales one and two (you do not put any coins on scale three)

A. If all three scales are balanced, all four sets of coins weight the same, and the remaining coin you have is the fake.

B. If two of three scales balance, all four sets of coins weigh the same, and the remaining coin you have is the fake. Also, the imbalanced scale is the fake one

C. If one of three scales balance, the balanced scale is an accurate scale (Note: it will be either scale one or two, it can't be scale three, the base)

D. If none of the scales balance, the third base scale is an accurate scale (The third scale must be accurate if both scales one and two are imbalanced, since at least one of the two actually has the heavy fake coin)

From here, you can simply divide the coins into groups of three and compare the weight on a gauranteed accurate scale. If this method is accepted as a single "weighing" then the answer can be obtained in only 5 measurements.

Note: I tried to be creative, but notice that the combination of three scales can also be done with three separate measurements instead if measuring them simultaneously like this is invalid. This leaves four measurements remaining, which is still enough to solve the problem since 3^4 = 81. The method of which has already been mentioned before and for brevity I won't repeat it here.

[Guest]

1. Using scale 1, compare 27 vs 27 coins.

2. Using scale 2, compare 12 vs 12 of the 27 coins indicated as heavy from the first weighing.

If = either scale 1 or scale 2 can still be either random or read true.

3. Using scale 2 again, compare the 3 not yet weighed from the 27 indicated as heavy from the first weighing vs three from the 24 from the second weighing.

If = either one of scales 1 or 2 is random and wrong so scale 3 must read true. Four more weighings using scale 3 reading true (27 vs 27, 9 vs 9, 3 vs 3, 1 vs 1) can determine the heavy coin.

If > or < then the first weighing must be accurate (tho scale 1 is either random or reads true) as at most only one of two scales can be wrong.

If < scale 2 is random and wrong as it contradicted itself so scales 1 and 3 must read true. Four more weighings using scale 1 or 3 reading true (27 vs 27, 9 vs 9, 3 vs 3, 1 vs 1) can determine the heavy coin.

4. Using scale 3, repeat the third weighing to confirm

If weighing 4 does not read the same as weighing 3, one of scales 2 or 3 is random and wrong. Three more weighings using scale 1 reading true (9 vs 9, 3 vs 3, 1 vs 1) can determine the heavy coin.

If weighing 4 and 3 are the same, the heavy coin is confirmed in the three indicated as heavy.

5. 6. 7. Using different scales compare 1 vs 1 of the three indicated as heavy.

Whichever reading occurs two or three times must be accurate and determines the heavy coin.

2. Using scale 2, compare 12 vs 12 of the 27 coins indicated as heavy from the first weighing.

If > or < the first weighing must be accurate (tho scale 1 is either random or reads true) as at most only one of the two scales can be wrong.

3. Using scale 3 compare 6 vs 6 of the 12 coins indicated as heavy from the second weighing.

If = then one of scales 2 or 3 is random and wrong so scale 1 must read true. Three more weighings using scale 1 reading true (9 vs 9, 3 vs 3, 1 vs 1) can determine the heavy coin.

If > or < the second weighing must be accurate as at most only one of the two scales can be wrong.

4. Using scale 1, compare 3 vs 3 of the 6 coins indicated as heavy from the third weighing.

If = then one of scales 3 or 1 is random and wrong so scale 2 must read true. Three more weighings using scale 2 reading true (6 vs 6, 3 vs 3, 1 vs 1) can determine the heavy coin.

If > or < the third weighing must be accurate as at most only one of the two scales can be wrong.

5. Using scale 2, compare 2 vs 1,X (known normal coin) of the 3 coins indicated as heavy from the fourth weighing.

If = then one of scales 2 or 1 is random and wrong so scale 3 must read true. Two more weighings using scale 3 reading true (3 vs 3, 1 vs 1) can determine the heavy coin.

If > or < the fourth weighing must be accurate as at most only one of the two scales can be wrong.

Label the three coins now known to contain the heavy coin a,b,c

5. Using scale 2 compare a,b vs c,X (already indicated as > or <)

6. Using scale 3 compare b,c vs a,X

7. Using scale 1 compare c,a vs b,X

If a is heavy, true readings of 5. 6. 7. would be >,<,>

If b is heavy, true readings of 5. 6. 7. would be >,>,<

If c is heavy, true readings of 5. 6. 7. would be <,>,>

Quick inspection will show that if any one of these readings is changed from a random scale reading incorrectly, the remaining two readings are still unique thus the heavy coin can be determined.

EDITS in red

Hope I didn't miss something. Easy to get confused with the random scale. A most excellent puzzle shakingdavid! Five stars from me.

Hi! Sorry for not replying sooner, I haven't been on here a lot lately and it takes some time to check a solution like this. Now I've read it and the only thing I don't understand with your solution is at the very end, weighings 5,6, and 7. Suppose you get the readings >,<,< which isn't among the three possible correct readings. How do you know if this is supposed to be >,<,> but scale 2 made weighing 5 incorrect, or >,>,< but scale 3 made weighing 6 incorrect?

[Guest]

I found the wrong step. Please cmment on my third modified solution in spoiler.

Well, after reviewing my earlier posts I came to know where I was wrong footed. Here is the modified solution.

Divide 81 coins in three groups 27A, 27B, & 27C.

W1S1:

Compare 27A & 27B in S1 balance. Three conditions are possible:

(a1) 27A = 27B; (b1) 27A > 27B; & (c1) 27A < 27B.

If we get (a1), then 27C would have heavier coin. So in next weighing in balance S2, if we get result which doesn’t show 27C side heavier, then it will be confirmed that one of the balance between S1 & S2 is broken i.e. S3 is OK. Then there will be no problem in finding the heavier coin within seven weighing.

So we will consider only the possibilities of getting those results in W2S2 which show 27C containing heavier coin.

W2S2:

Divide 27B & 27C each in two groups - 14B, 13B & 14C & 13C. Compare them in S2. Three conditions are possible:

(b2) 14B + 13C > 13B + 14C; (c2) 14B + 13C < 13B + 14C.

If we get result (b2), then 13C would have heavier coin; so in next weighing W3S3, if we get result which does not show 13C side heavier, then it will be confirmed that one of the balance between S2 & S3 is broken i.e. S1 is OK. Then there will be no problem in finding the heavier coin within seven weighing.

So we will consider only the possibilities of getting those results in W3S3 which show 13C containing heavier coin.

W3S3:

Divide 13C in groups 7C, 6C and 13B in 7B, 6B. Comparing them in S3 may give one of the following results:

(b3) 7C + 6B > 7B + 6C; (c3) 7C + 6B < 7B + 6C;

If we get result (b3), then 7C would have heavier coin; so in next weighing W4S1, if we get result which does not show 7C side heavier, then it will be confirmed that one of the balance between S1 & S3 is broken i.e. S2 is OK. Then there will be no problem in finding the heavier coin within seven weighing.

So we will consider only the possibilities of getting those results in W4S1 which show 7C containing heavier coin.

W4S1:

Divide 7C in groups 4C, 3C and 13B in 4B, 3B. Comparing them in S1 may give one of the following results:

(b4) 4C + 3B > 4B + 3C; (c4) 4C + 3B < 4B + 3C;

If we get result (b4), then 4C would have heavier coin; so in next weighing W5S2, if we get result which does not show 4C side heavier, then it will be confirmed that one of the balance between S1 & S3 is broken i.e. S2 is OK. Then there will be no problem in finding the heavier coin within seven weighing.

So we will consider only the possibilities of getting those results in W5S2 which show 4C containing heavier coin.

W5S2:

Divide 4C in groups 2C1, 2C2 and 4B in 2B1, 2B2. Comparing them in S2 may give one of the following results:

(b5) 2C1 + 2B1 > 2B2 + 2C2; (c5) 2C1 + 2B1 < 2B2 + 2C2;

If we get result (b5), then 2C1 would have heavier coin; so in next weighing W6S3, if we get result which does not show 2C1 side heavier, then it will be confirmed that one of the balance between S2 & S3 is broken i.e. S1 is OK. Then there will be no problem in finding the heavier coin within seven weighing.

So we will consider only the possibilities of getting those results in W6S3 which show 2C1 containing heavier coin.

W6S3:

Compare two coins of 2C1 in S3 balance, one of the coin will show heavier. Heavier coin may be confirmed in W7S1.

Same procedure will follow for other results (b1), (c1) also only difference will be that we will have to take 27A or 27B whichever heavier, in place of 27C.

Hi! This is an interesting approach that you and plainglazed have taken, but I'm not sure if it works. Your solution works until W5S2, but doesn't work at the end. If W5S2 gives weighing (b5) indicating that 2C1 has the heavy coin, but W6S3 gives equality indicating that 2C1 does not have the heavy coin, then you are right that one of scales 2 and 3 is random, so scale 1 is OK. But now you have only one weighing left and four possible coins, so you can't solve it.