ES's Mathematical Puzzle - 1

I think it will need 4 try to determine the coin, either heavy or light. But if it is fixed that coin is either heavy or light than it can be decided in 3 try.

Here is my method for solving the puzzle :
1-> Take any 6 coins & divide it in two parts. Weigh this 3-3 coins, if the scale is equal than the remaining 6 coins contains the odd coin. If not than this six coin cotains the odd coin.
Now we have 6 coins of equal weight say group A & another 6 coins containing the odd coin say group B.
2-> Now take 3-3 coins from both A & B, and measure it, if the scale is equal than the remaing 3 coins contains the odd coin, if not then this 3 coins of group B contains the odd coin.
Now we have 9 coins of equal weight in group A, and 3 coins containing one odd coin in group B.
3-> Now measure 2 coins of group B with 2 coins of group A. If the scale is equal than we have no problem, cause the remaining coin will be the odd coin. But if not than we need one more to determine the odd coin.
So now we have 10 coins in group A and 2 coins in group B if scale is not equal.
4-> so in last try compare 1 coin of group A with 1 coin of group B, if it is epual than the remaining coin is odd coin, if not than this coin is odd coin.

If we know that coin is either heavy or light than the we can determine the odd coin in 3 tries. Here is the solution for this by assuming the coin is heavy ( or light ).
-> repeat step 1 & 2 as above.
3-> take any 2 coins from group B, if the scale is equal than the remaining coin is odd coin. If not than, the heavy ( or light ) coin will be the odd coin.

Now ES, let me know whether I am right or wrong.

As per me the answer is fine but i am not sure though 😀 need to check and investigate i too came to the same solution so as perme its correct
Congrats Rajdeep 😀

Yep. I too got it in 4 uses of the balance.

Can any one do it with three uses??

No..............
3 chances are sufficient. I have solved this but it's damn hard to explain it for me, here. I will get back !

To resolve 12 coins, divide them into a two groups. One of 9 coins (G1) and another group of 3 coins (G2). Further divide these groups G1 and G2 into three equal subgroups A, B, C, of 3 coins each and three equal groups a, b, c of one coin each. Now STart weighing in the following fashion:

1. Left(A and a) against Right(B and b).
2. Left(B and a) against Right(C and b).

3. There are two cases for the third weighing,

Case 1: according to the results of the first two weighings. If both results are the same (means left pane is lighter/heavier then right pane in both the weighing), then the false coin is in the G2 group, and we already know from the first two weighings the result of a against b (which is lighter/heavier). Therefore our third weighing is between b and c.

Case:2 If the two results differ(result 1 and 2), then the false coin is in the G1 group, and we know which of A, B, C it is in, and whether it is heavy or light. The false subgroup contains three coins, so balance any two against each other. If they balance, the false coin is the third one, and if they don't balance, the false coin is the heavy or light side according to whether the false coin is know to be heavy or light.

-CB

crazyboyTo resolve 12 coins, divide them into a two groups. One of 9 coins (G1) and another group of 3 coins (G2). Further divide these groups into three equal subgroups A, B, C, of 3 coins each and three equal groups a, b, c of one coin each. Now STart weighing in the following fashion:

1. Left(A and a) against Right(B and b).
2. Left(B and a) against Right(C and b).

3. There are two cases for the third weighing,

Case 1: according to the results of the first two weighings. If both results are the same (means left pane is lighter/heavier then right pane in both the weighing), then the false coin is in the G2 group, and we already know from the first two weighings the result of a against b (which is lighter/heavier). Therefore our third weighing is between b and c.

Case:2 If the two results differ(result 1 and 2), then the false coin is in the G1 group, and we know which of A, B, C it is in, and whether it is heavy or light. The false subgroup contains three coins, so balance any two against each other. If they balance, the false coin is the third one, and if they don't balance, the false coin is the heavy or light side according to whether the false coin is know to be heavy or light.

-CB

I'm sorry. I can't get your solution. Can you explain it a bit more elaborate?? Thanks!! 😀

silverscorpionI'm sorry. I can't get your solution. Can you explain it a bit more elaborate?? Thanks!! 😀

At which point you lost, I will explain from their onwards.

-CB

I am not getting your solution from case 1. And one more thing, as I got through your solution, I come to know that, assume the coin 'a' of group G2 is the odd coin, and if we go through your solution, then it cant be determined. Here is the explanation of your solution as per me with assumin 'a' is the coin which we want to find :
upto 2 tries it is right. Then if a is either lighter or heavier then try 1 will not give the equal scale and try 2 will give the equal reading. So we have to go to the case 2, according to your solution. So the odd coin is in group G1, which cant be possible. May be you should think about me. If I am wrong in this explanation than I am sorry for that.

Here is my explanation.

Take 4 coins in one pan of weight balance and 4 in another pan and compare them using first chance of using weight balance. Keep 4 coins aside.

Case 1: The weight balance shows equilibrium
Case 2: The weight balance shows difference

Case 1: The weight balance shows equilibrium. This means none of these 8 coins is fake. Now take any 3 of the remaining 4 coins in one pan(say left pan) and 3 good coins in other pan(right pan). Keep one coin aside. Compare using second chance of using weight balance.
Case 1A: The weight balance shows equilibrium
Case 1B: The left pan is lighter
Case 1c: The left pan is heavier

Case 1A: Means the one coin which was kept aside is fake. Compare it with any one of good coins using last chance to decide if it's lighter or heavier.

Case 1B: The left pan is lighter: Means one of these 3 coins is fake and is lighter. Compare any 2 of these lighter coins aginst each other using third chance of using weight balance. If balance shows equilibrium, third coin (which was kept aside in last comparison) is fake and oviously lighter. If balance shows difference, the lighter pan has fake coin. (because we know it is lighter)

Case 1C: The left pan is heavier: This case is exactly same as above case(1B) except fake coin is heavier here.

Case 2 : The weight balance of first comparison shows difference. Now, mark 4 coins from lighter pan as "L" and 4 coins from heavier pan as "H". This indicates if fake coin is in L group it MUST BE lighter and if it is in H group it MUST BE heavier. Now take two L coins and two H coins in left pan whereas one L coin, one H coin and two good coins in right pan. Keep 1 L and 1 H coin aside. Compare using second chance of weight balance.

Case 2A: The weight balance shows equilibrium
Case 2B: The left pan is lighter
Case 2c: The left pan is heavier

Case 2A: The weight balance shows equilibrium: Means eaither of the L and H coins left aside is fake. Compare any of them against good coin using third chance to determine which one is fake. If H is fake it is heavier and fake. If L coin is fake it is lighter and fake.

Case 2B: The left pan is lighter: Now, left pan can be lighter if i) any of two L coins in left pan is lighter OR ii) H coin in right pan is heavier.
Now, compare these 2 L coins from left pan against each other using third chance. If equilibrium, H coin is fake (needless to say it is heavier). If difference, lighter pan has fake coin and obviously it is lighter.

Case 2C: The left pan is heavier: This case is similar to above case. Now, left pan can be heavier if i) any of two H coins in left pan is heavier OR ii) L coin in right pan is lighter. Now, compare these 2 H coins from left pan against each other using third chance. If equilibrium, L coins is fake (needless to say it is lighter). If difference, heavier pan has fake coin and
obviously it is heavier.


Hussssssssh! Am I right?

Brilliant!!!!!
I just leave these puzzle after messing with my mind too much.
No need to say again, but you are really genious.
Hats off to you!!!!

RajdeepCEI am not getting your solution from case 1. And one more thing, as I got through your solution, I come to know that, assume the coin 'a' of group G2 is the odd coin, and if we go through your solution, then it cant be determined. Here is the explanation of your solution as per me with assumin 'a' is the coin which we want to find :
upto 2 tries it is right. Then if a is either lighter or heavier then try 1 will not give the equal scale and try 2 will give the equal reading. So we have to go to the case 2, according to your solution. So the odd coin is in group G1, which cant be possible. May be you should think about me. If I am wrong in this explanation than I am sorry for that.

You are getting it absolutely wrong 😁

Ok, Lets take your example only. Suppose faulty coin is in group G2 (suppose a). So first two weighings are :

1. Left(A and a) against Right(B and b).
2. Left(B and a) against Right(C and b).

Now If a is faulty that means all 3 coins in A, B, C are correct. So I can theoritically replace A with B with C. Noe equation would be:
A+a = A+b
A+a = A+b (replaced C\B with A as all three coins in A, B and C are same)

So now I should go in case I. I guess it will help you in understanding...

-CB

Sorry crazyboy, cause I just forgot at one point. Anyway your solution is also right, but it will also need 4 tries to determine the fake coin if we dont know the weight. Am I right?

RajdeepCEBrilliant!!!!!
I just leave these puzzle after messing with my mind too much.
No need to say again, but you are really genious.
Hats off to you!!!!

Thanks a lot for those kind words! 😀 But I think, too much appriciation for a old guy like me ! 😀

--Diff

RajdeepCESorry crazyboy, cause I just forgot at one point. Anyway your solution is also right, but it will also need 4 tries to determine the fake coin if we dont know the weight. Am I right?

No buddy it will takes 3 tries only... please re-read and a bot more concetration... 😀

-CB

Yes you are right crazyboy. I just lost at case 2, but this time I carefully follow your solution. Your explanation is complicated so it takes time to understanding me.

ES - next puzzle please?

The point I don't get is, if the results of both the weighings are same, how do you say the odd coin is in the group B? Similarly for group A??

And also, how do you determine if the odd coin is lighter or heavier??

oh, sorry. I didn't see the other posts.
Seems it's already explained..

Now I understand it. Thanks CB.
And differential, yours is also brilliant. Keep it up!! 😀

And differential, yours is also brilliant. Keep it up!! :smile:

Thanks buddy for those sweet words ! 😀

--Diff

Excellent answers guys.

@CB:: Next puzzle will be posted to night .