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]

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!

this would work great if you had a 3 pan balance, and used the 3rd pan for a known weight, e.g. 130g.

Without having a weight reading is the real challenge.

[Guest]

my answer is 5 times.. heres how it goes

27 coins = 8 8 8 1 1 1. thats a b c d e f

if a = b, a = c, we go to the ones.. if d = e, then the answer is F. COUNT = 3

if a = b, a = c, we go to the ones.. if d =/ e, we weight f. if d =/ f then the answer is D. COUNT = 4

if a = b, a = c, d =/ e, d = f, the answer is E. COUNT = 4

now we go to the if the diff coin is at a or b or c.

if a = b, a =/ c, then we measure c.

before we go further, i want to show you how we solve 8 coins with 3 scaling.

8 coins we devide it into twos.

8 coins = 2 2 2 2, thats w x y z.

if w = x, w = y, then we weight 1 coin from w, and 1 coin from z. if w = z, then the other coin from z is the answer.

if w = x, w =/ y, then we weight 1 coin from w, and 1 coin from y. if w =/ y, the answer is that coin from y.

if w =/ x, w = y, then we weight 1 coin from w, and 1 coin from x. if w = x, the other coin from x is the asnwer.

if w =/ x, w =/ y, one coin from W is the answer. we weight 1 coin from W, 1 coin from x or y or z..

TOTAL COUNT = 3

if a = b, then the coin is at C. total count 4.

if a =/ b, a =/ c, then the coin is at A. total count 5.

if a =/ b, a = c, then the coin is at B. total count 5.

Edited October 2, 2009 by kyky

[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?

look at my answer =) hope helps.

[Guest]

Yes... I seem to find this problem a tricky one... I heard it another way (the one coin is heavier). If the one coin was heavier, then the answer would have been

3

But because your coin is either heavier or lighter and you dont know, you have to check it

one extra time

I love this one!

weigh 9 against 9 and take one pile and weigh that against the other 9 that wasnt used. THIS will give you the information you need (the coin is lighter or heavier?). Then you can solve.

I know I am kind of late on answering this... but I just wanted to go back to it. FACE PALM!

JUST CLARIFICATION:

9 to 9, then check with another. Example: 9 to 9 and they are different. Take heavier one and weigh against the 9 not used. It is the same as this one. The coin is in the light group in the first weighing. THEN: Take that light group and separate into 3 piles and weigh 3 - 3, and say they are light and heavy, take the light one and go 1-1, take the light one. IN THIS EXAMPLE, TAKE THE LIGHT ONE! (Note, this has no proof (At least in my post) so DO NOT REFER TO MY POST IF YOU WANT TO DO SOMETHING WITH THIS PUZZLE!)

I hope this post helped at all.

Edited October 2, 2009 by Nikk29

[Guest]

look at my answer =) hope helps.

Note, this problem has been solved earlier in the postings. The answer is not what you got, but you did great answering it. You can read another person's post, and hopefully it helps you understand. You DID do well in explaining your answer though.

[Guest]

Note, this problem has been solved earlier in the postings. The answer is not what you got, but you did great answering it. You can read another person's post, and hopefully it helps you understand. You DID do well in explaining your answer though.

thanks for the compliment, but no i dont read others answer before i got mine.

[Guest]

5

Edited October 27, 2009 by varin

[Guest]

The answer is 5

The answer is 5

first make three set of 9 coins say set 1, set 2, set 3

step 1. measure set1 and set2

three posibilities set1>set2 or set 2>set1 or set 1= set2

in case set 1= set2 go to 3rd step

step 2 compare set 1 and set 3

three conditions set1>set3 or set set 1< set3 or set 1=set3

if set1 =set 3 then the coin is in set 2 and if set 2>set3 that means the coin is heavier i.e. it is of 11g and if it is less it is of 9 g

if set1>set3or set 1<set3 then the coin is in set 3 and if set 3>set1 that means the coin is heavier i.e. it is of 11g and if it is less it is of 9 g

step 3 now you have 9 coins divide these 9 into 3 set of set 1, set 2, set3 repeat step 1

step 4 repeat step 2

step 5 now you have 3 coins and you know the coin is heavier or lesser

if coin is heavier measure coins 1 and 2 first now three things

1>2 or 2>1 or 1=2 if 1>2 then the odd one is coin 1 and if 2>1 then the odd coin is 2 and if they are equal then odd coin is 3

similarly if coin is lesser measure coins 1 and 2 first now three things

1>2 or 2>1 or 1=2 if 1>2 then the odd one is coin 2 and if 2>1 then the odd coin is 1 and if they are equal then odd coin is 3

first make three set of 9 coins say set 1, set 2, set 3

step 1. measure set1 and set2

three posibilities set1>set2 or set 2>set1 or set 1= set2

in case set 1= set2 go to 3rd step

step 2 compare set 1 and set 3

three conditions set1>set3 or set set 1< set3 or set 1=set3

if set1 =set 3 then the coin is in set 2 and if set 2>set3 that means the coin is heavier i.e. it is of 11g and if it is less it is of 9 g

if set1>set3or set 1<set3 then the coin is in set 3 and if set 3>set1 that means the coin is heavier i.e. it is of 11g and if it is less it is of 9 g

step 3 now you have 9 coins divide these 9 into 3 set of set 1, set 2, set3 repeat step 1

step 4 repeat step 2

step 5 now you have 3 coins and you know the coin is heavier or lesser

if coin is heavier measure coins 1 and 2 first now three things

1>2 or 2>1 or 1=2 if 1>2 then the odd one is coin 1 and if 2>1 then the odd coin is 2 and if they are equal then odd coin is 3

similarly if coin is lesser measure coins 1 and 2 first now three things

1>2 or 2>1 or 1=2 if 1>2 then the odd one is coin 2 and if 2>1 then the odd coin is 1 and if they are equal then odd coin is 3

[Guest]

In minimum 6 attempts one can get the definite answer.

Make 27 as 9,9,9 we say (pile1,pile2,pile3)

1.Weigh 2 piles (pile 1 & pile 2), If they are balanced then the coin is definitely in 3rd pile (pile 3) .

2.If not then again weigh either of the pile(pile 1) with the remaining pile (pile 3).

(If both sides are balanced the different coin is in the pile 2, If not its in pile 1)

Now we are having 9 coins, again make 3 piles of 3 coins in each

3.Repeat step 1 with new 3 piles (pile 4,pile 5, pile 6)

4.Repeat step 2

Now we are having 3 coins,

5.Weigh 2 coins (coin 1 & coin 2), If they are balanced then the 3rd coin is different.

6.If not then again weigh either of the coin (coin 1) with the remaining coin (coin 3).

(If both sides are balanced the different coin is coin 2, If not its coin 1)

Edited January 14, 2010 by satanite

[Guest]

"Parik" posted on 18 March 2009 - 04:46 PM the clearest description I have seen to solving the case comprehensively.

That is by dividing into groups of 9, then 3 and finally singly.

In each Division we can identify the defective group or item in 1 or 2 steps, depending on luck.

Using this method, the best outcome is 3 weighings and the worst outcome is 6 weighings. I think this is the best strategy to adopt as the arithmetic average of the number of attempts is 4.5.

Other strategies like balancing 13-13 and leaving one aside may solve the problem in 1 step (very lucky) or end up with 26 steps (if by 26 attempts the odd coin has not been identified, we can infer that the last coin is the odd one without further weighing). The arithmetic average of all possible outcomes is 13.5.

[Guest]

For finding the different one out of three u will always require just 2 attempts..

suppose u try with 1 and 2 bunch they are not equal.

try with 1 and 3 and now if they are also not equal then its sure that the bunch 1 contains fault else if 1 and 3 are equal then bunch 2 will have fault in it.

So just 2 try for a lot of 3.

Thus in just 6 attempts you could find the solution.

So 6 is the answer..

[Guest]

In min 3 weighings and max 5 weighings we can definitely find the odd coin

All possibilities :-

ivide 27 into piles of 7, 7, 13

1) compare 7 vs 7 coins, keeping 13 coins aside

If both of them don't weigh same, then it means that the odd coin is inside these 14 coins

If both of the sets weigh same, then we need to find the odd coin from 13 coins.

2) compare 4 vs 4 coins, keeping 5 or 6 coins aside(this depends on the previous result)

Applying the same logic as explained in step 1, we will be left with 5 or 6 or 8 coins to find the odd coin from

3) Depending on the result from previous step we need to split coins to weigh

case 1) 5 - 2vs2, 1 kept aside

case 2) 6 - 2vs2, 2 kept aside

case 3) 8 - 3vs3, 2 kept aside

If its case one, then we are done, else we will be left with either 2 or 3 coins to determine the odd coin from

4) If two coins are left, then we can find the odd one by comparing with normal coins(other coins)

If three coins are left, then we may require one more weighing

[Guest]

An easier-to-understand (IMHO): (sorry for my bad english)

27 ==> 13[A] - 13 - 1[C]

compare A and B (1 weighing)

if A=B then we found the answer in C

13 ==> 6[A] - 6 - 1[C] --

compare A and B (1 weighing)

if A=B then we found the answer in C

6 ==> 3[A] - 3

compare A and B (1 weighing)

then we check either A or B is heavier

3 ==> 1[A] - 1 - 1[C]

compare A and B (1 weighing)

if A=B then we found the answer in C

Then, I can assure one to weigh at least 4 times before I get the correct answer.

For you who are studying Computer Science, you might notice that this solution uses

Divide-and-Conquer approach.

Edited July 13, 2010 by Teddy Kesgar

[Guest]

An easier-to-understand (IMHO): (sorry for my bad english)

27 ==> 13[A] - 13 - 1[C]

compare A and B (1 weighing)

if A=B then we found the answer in C

13 ==> 6[A] - 6 - 1[C] --

compare A and B (1 weighing)

if A=B then we found the answer in C

6 ==> 3[A] - 3

compare A and B (1 weighing)

then we check either A or B is heavier

3 ==> 1[A] - 1 - 1[C]

compare A and B (1 weighing)

if A=B then we found the answer in C

Then, I can assure one to weigh at least 4 times before I get the correct answer.

For you who are studying Computer Science, you might notice that this solution uses

Divide-and-Conquer approach.

which one among A or B will you choose for ur next step as it says one of them is either heavy or light ... So u cannot determine the defective bunch this way ..

[Guest]

27=1+13+13

balance 13coins nd 13 coins... if equal ... the one left is the heavy/low weight coin..

otherwise,

13=1+6+6

continue the same process......

6=1+3+3

continue...

3=1+1+1

continue....

minimum steps is < 4...

[Guest]

make 3 group of 9s, balance any two groups to observe: (a) it's a balance then third group has faulty coin --- 1 weighing, anomaly unknown (b) it's tilted on one side, in which case replace the heavier group with the third group to see (b.i) the balance --- 2 weighing, anomaly is 11g coin in heavier group (b.ii) similar tilt --- 2 weighing, anomaly is 9g coin in lighter group repeat the above steps by diving the identified group in 3 subgroup of 3s, and then further into 3 subgroups of 1s. In short 3 weighing to identify the coin and 1 to identify the anomaly.

overall 4 weighing...

[Guest]

Using your method you need only 6 weighing .. use a balance scale not a numerical one . with 3 piles you only need to weigh twice, to find the odd one out. weigh pile 1 against 2,. if equal then pile 3 is the odd one . if unequal then weigh the heavier against pile 3 .. this will be equal or lighter . if unequal and it cannot be heavier . that would mean 3 different weights, so pile 3 is the odd ball. do this with the next 2 sets . equaling 6 at most.

[Guest]

The answer is 4 and my solution is below.

if(b1:b9 == b10:b18){

if(b19:b21 == b22:b24){

if(b25 == b26)

{

***b27***

}else//if(b25 == b26)

{

if(b25 == b27)

{

***b26***

}

else

{

***b25***

}

}

}

else //if(b19:b21 == b22:b24)

{

case(b19:b21 < b22:b24){

if(b19,b22,b21 < b20,b26,b27){ //b23,b24 left out

//that means b19 b 21 is lighter

if(b19<b21)

{

***b19***

}

else

{

***b21***

}

}else if(b19,b22,b21 > b20,b26,b27)

{ // that means b20 is lighter

***b20***

}

else if(b19,b22,b21 == b20,b26,b27)

{ // that means left out balls where heavier

if(b23 < b24)

{

***b24***

}

else

{

***b23***

}

}

}

case(b19:b21 > b22:b24){

// same approach as case(b19:b21 < b22:b24)

}

}

}

else //if(b1:b9 == b10:b18)

{

case(b1:b9 < b10:b18){

if(b1,b2,b3,b10,b11,b12,b7,b8,b9 < b4,b5,b6 + 6 equal balls){ //b13:b18 left out

// this means b1,b2,b3,b7,b8,b9 might be ligther

if(b1:b3 < b7:b9){

if(b1 <b2)

{

***b1***

}

else if(b1 > b2)

{

***b2***

}

else(b1 == b2)

{

***b3***

}

}

else

{

// same solutiın as above

}

}else if (b1,b2,b3,b10,b11,b12,b7,b8,b9 > b4,b5,b6 + 6 equal balls){

// this means b10:b12 balls are heavier

if(b10 > b11){

***b10***

}

else if ( b10 < b11){

***b11***

}

else{

***b12***

}

}else if (b1,b2,b3,b10,b11,b12,b7,b8,b9 == b4,b5,b6 + 6 equal balls){

// this means left out balls b13:b18 are heavier

if(b13:b15 > b16:b18){

if(b13 > b14){

***b13***

}

if(b13 < b14){

***b14***

}

if(b13 == b14){

***b15***

}

}

else

{

// same solution as above

}

}

}

case(b1:b9 > b10:b18){

// same solution as above

}

}

[Guest]

we can get the answer by 4 measurements exactly.

divide them three groups.. A B C -- will have 9 coins each.

step 1:

A, B

step 2:

A, C

step 3: -- got x

xA, xB -- xA, xB, xC will have 3 coins each, x is answer group from steps 1,2.

step 4: -- got xx

xxA, xxB -- xxA, xxB, xxC will have 1 coin each, xx is answer group from step 3.

steps 1 and 2:

--------------

if (A > B & A > C) then

take group A as x -- odd coin is heavier

else if (A < B & A < C) then

take group A as x -- odd coin is lighter

else if (A < B & A = C) then

take group B as x -- odd coin is heavier

else if (A > B & A = C) then

take group B as x -- odd coin is lighter

else if (A = B & A < C) then

take group C as x -- odd coin is heavier

else if (A = B & A < C) then

take group C as x -- odd coin is lighter

end

step 3:

--------- by knowing coin is lighter/heavier continue this step

-- consider it is heavier

if (xA > xB) then

take group xA as xx

else if (xA < xB) then

take group xB as xx

else if (xA = xB) then

take group xC as xx

end

-- consider it is lighter

if (xA < xB) then

take group xA as xx

else if (xA > xB) then

take group xB as xx

else if (xA = xB) then

take group xC as xx

end

step 4:

-------

-- consider it is heavier

if (xxA > xxB) then

heavier coin is xxA

else if (xxA < xxB) then

heavier coin is xxB

else if (xxA = xxB) then

heavier coin is xxC

end

-- consider it is lighter

if (xxA < xxB) then

lighter coin is xxA

else if (xxA > xxB) then

lighter coin is xxB

else if (xxA = xxB) then

lighter coin is xxC

end

[Guest]

THE ANSWER IS SIX

When the coins are divided in three groups, two measures are enough to decide which group is different.

Thus divide the coins in three groups of 9 coins and measure them twice.

Then divide the different group in three groups of 3 and measure them twice again.

Then divide the different group in three coins and measure them twice.

TOTAL = SIX

[Guest]

THE ANSWER IS SIX

When the coins are divided in three groups, two measures are enough to decide which group is different.

Thus divide the coins in three groups of 9 coins and measure them twice.

Then divide the different group in three groups of 3 and measure them twice again.

Then divide the different group in three coins and measure them twice.

TOTAL = SIX

that was my initial answer too (3*2, measuring it twice after each time you divide by 3)

but I realized that after dividing 27 into three 9-coin groups and measuring twice, you'll know whether the odd coin out is heavier or lighter, after that you'll only need to measure once after each time you divide by 3

6 is the answer if the weighing device only tells you whether two groups of coins are the "same" or "different" in weight (no weighing device that I know of does only that)

if the weighing device tells you which group of coins is heavier and which is lighter, the answer is 4

Edited March 16, 2011 by b1soul

[Guest]

We can do it in 3 Measurements

Make 3 Groups of 9 Coins Each

Measure1 : Weigh Any Two Groups , If they weigh same . 3rd which is kept aside contains the required coin else one with heavy side has the required coin.. This Eliminates 18 Coins out of 27

Measure2: Elliminates 6 Coins Going by Above Process . Balance is now 3

Measure3: Weigh Any 2 , if same the coin kept aside is required coin else the one on heavy side

[Guest]

The answer is not 3 as suggested above because you don't know if the different coin is heavier or lighter. With 3 attempts you can only do this if there are 12 coins. This can be done in 4 attempts. Actually using 4 attempts, you can find the different coin and know if it is heavier or lighter given 39 coins (not just 27). I will try to explain:

We have 3 groups of 13 coins. Compare 2 groups of 13. (measure #1)

if even, the different one is in the 13 that was not compared.

...Take 9 of those and compare them against 9 coins that you know are the same (measure #2)

...if even, then the different one is one of the remain 4.

......Take 3 of those and compare them against 3 coins that you know are the same (measure #3)

......if the same, then the different one is the one not measured

.........Compare the last coin against any other coin and see if it heavier or lighter (measure #4)

......if not the same, then it is one of the 3, and we know if it is heavier or lighter

.........take 2 of them and compare them against one another (measure #4)

.........if the same then it is the coin that we didn't measure and we already know if it is heavier or lighter

.........if not the same, since we already know the different one is heavier or lighter, then that is the one

...if not even, then it is one of the 9, and we know if it is heavier or lighter

......divide these 9 into three groups of three and compare 2 of the groups (measure #3)

......if the same, then it is one of the three that we didn't measure

.........take 2 of them and compare them against one another (measure #4)

.........if the same then it is the coin that we didn't measure and we already know if it is heavier or lighter

.........if not the same, since we already know the different one is heavier or lighter, then that is the one

......if not the same, then since we already know that the different one is heavier or lighter,

......then we know which group of the 3 it belongs to

.........take 2 of them and compare them against one another (measure #4)

.........if the same then it is the coin that we didn't measure and we already know if it is heavier or lighter

.........if not the same, since we already know the different one is heavier or lighter, then that is the one

if not even, then the 13 the we put aside are the same, and so then take out 9 of the coins from the one side, say the heavier side, move 9 coins from the lighter side to the heavier side, and add 9 of the coins from the pile that we didn't measure to the lighter side, and measure again (measure #2)

...if the same, then it is one of the 9 coins that we took out, and we know the different one is heavier

......divide these 9 into three groups of three and compare 2 of the groups (measure #3)

......if the same, then it is one of the three that we didn't measure

.........take 2 of them and compare them against one another (measure #4)

.........if the same then it is the coin that we didn't measure and we already know it is heavier

.........if not the same, then it is the heavier one

......if not the same, then we know it is on the heavier side

.........take 2 of them and compare them against one another (measure #4)

.........if the same then it is the coin that we didn't measure and we already know it is heavier

.........if not the same, then it is the heavier one

...if the scale changed direction, then it is one of the 9 coins that we moved from the lighter side,

...and we know the different one is lighter

......divide these 9 into three groups of three and compare 2 of the groups (measure #3)

......if the same, then it is one of the three that we didn't measure

.........take 2 of them and compare them against one another (measure #4)

.........if the same then it is the coin that we didn't measure and we already know it is lighter

.........if not the same, then it is the lighter one

......if not the same, then we know it is on the lighter side

.........take 2 of them and compare them against one another (measure #4)

.........if the same then it is the coin that we didn't measure and we already know it is lighter

.........if not the same, then it is the lighter one

...if the scale remained the same, then it is one of the 8 coins (4 on each side) that we didn't move

......take out 3 of the coins from the heavier side, move 3 coins from the lighter side to the heavier side,

......and add 3 of the coins that we know are the same to the lighter side, and measure again (measure #3)

......if the same, then it is one of the 3 coins that we took out, and we know the different one is heavier

.........take 2 of them and compare them against one another (measure #4)

.........if the same then it is the coin that we didn't measure and we already know it is heavier

.........if not the same, then it is the heavier one

......if the scale changed direction, then it is one of the 3 coins that we moved from the lighter side,

......and we know the different one is lighter

.........take 2 of them and compare them against one another (measure #4)

.........if the same then it is the coin that we didn't measure and we already know it is lighter

.........if not the same, then it is the lighter one

......if the scale remained the same, then it is one of the 2 coins (1 on each side) that we didn't move.

......Compare the heavier one against one coin that we know is the same as other (measure #4)

.........if the same then it is the one that we didn't measure and it is lighter

.........if not the same then it is the one that we just measure and it is heavier

I hope this was clear enough

[bhramarraj]

Divide 27 coins in 3 groups, each of 9 coins, say 9A, 9B, & 9C.

W1: Compare 9A & 9B. Possible results are:

[1] 9A = 9B, OR

[2] 9A > 9B, OR

[3] 9A < 9B.

W1 [1]: If 9A = 9B, then divide 9C in 3 groups of 3 coins each, say 3C1, 3C2, & 3C3.

w2: Compare 3C1 & 3C2. If 3C1 >or < 3C2, then

w3: Compare 3C1 & 3C3, if 3C1 = 3C3, then counterfeit coin is in 3C2 and is lighter if 3C1 > 3C2 and heavier if 3C1 < 3C2. So w4 will tell which coin in 3C2 is lighter or heavier. And if 3C1 > or < 3C3 in w3, then you know counterfeit coin is in 3C2 and whether it is heavier or lighter. So w4 will tell the required coin.

But If w2 gives 3C1 =3C2, then counterfeit coin is in 3C3. Divide 3C3 in 3 groups of 1 coin, say 1C4, 1C5, & 1C6.

w3: Compare 1C4 & 1C5. If you get 1C4 > or < 1C5, then compare 1C4 & 1C6 in w4, which will decide the counterfeit coin and whether it is heavier or lighter.

But if you get 1C4 = 1C5, in w3, then counterfeit coin is 1C6.

So compare 1C4 & 1C6 to find whether 1C6 is heavier or lighter in w4.

------------------------------------------------------------------------------------------------------------

W1 [2]: If W1 shows 9A > 9B, then either 9A has heavier or 9Bhas lighter counterfeit coin.

Divide 9A in three groups 3A1, 3A2, & 3A3.

W2: Compare 3A1 & 3A2. Possible results are: [4]3A1 = 3A2; [5]3A1 > 3A2; [6] 3A1 < 3A2

W2 [5] or W2 [6]: If 3A1 > 3A2, then the counterfeit coin is heavier and is in group 3A1 because 9A is heavier than 9B. Similarly if 3A1 < 3A2, than heavier coin is in 3A2. In both cases, the third weighing will tell the heavier coin.

W2 [4]: If 3A1 = 3A2, then counterfeit coin may be heavier in 3A3, or lighter in 9B.

So divide 9B in three groups 3B1, 3B2, & 3B3.

W 3: Compare 3B1 & 3B2, to get either of the following results:

[7] 3B1 = 3B2; [8] 3B1 > 3B2; [9] 3B1 < 3B2.

If we get W3 [7]: Counterfeit coin may be either heavier in 3A3 or lighter in 3B3.

W4: Compare 3A3 with 3A2, if both are equal, then the coin is lighter and is in group 3B3. But if 3A3 > 3A2, then you know that counterfeit coin is heavier and is in 3A3 group. In both cases you can find out the counterfeit coin in w5, and as you know whether it is heavier or lighter.

But if we get results [8] or [9] from w3, and you know the counterfeit coin is lighter (because 9B is lighter group), therefore you know that in case [8] it is the group 3B2, and in case [9] it is the group 3B1 which contains the lighter coin. So in w4 you can separate out the counterfeit coin.

You can find out the counterfeit coin and whether lighter or heavier in maximum 5weighing, by adopting the same procedure if you get result [3] 9A < 9B in W1.

[Guest]

if you are not lucky , u will solve it in at max 6 tries...if u r somewhat lucky,u can do it in 5..and u can do it in 4 if u r very lucky...

IF U UNDERSTAND THIS IT MEANS YOU HAVE THE COMPLETE ANSWER.