Weighing

[smart_coder]

There are 12 gold coins of same material, weight , colour and everything except one gold coin which is different in weight from other 11 coins.
U hve been given a weighing balance and u are allowed to do only 3 weighings. How will u find the defected gold coin ?

[psuresh1982]

Solution 1:
First divide the 12 coins into (5,5,2)

Weight1: First compare two 5 coins. If both of them equal then goto weight3. Otherwise divide the weightless set coins into (2,2,1)

Weight2: Second compare the two 2 coins. If the two weights are equal then the remaining one is defected coin. Othewise goto the weight3

Weight3: Now you have only two coins. Weight this and found the defected coin.


Solution2:
First divide the 12 coins into (2,2,8)

Weight1: First compare two 2 coins. If both of them equal then divide the 8 coins into (3,3,2). If not then take weighless two coins and then weight again. After this you got a defected gold coin.

Weight2: Second compare the two 3 coins. If the two weights are equal then the remaining two coins contains the defected coin. Otherwise you got a weightless 3 set of coins.

Weight3: From the previous weight either you got 3 set coins or 2 set coins. If you got three set coins then weight any other two and check it. If its equal then the remaining one is defected. Otherwise you can found defected coin using the weights. Now if you have 2 set coins then weight the two coins and found the defected coin.


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suresh

[smart_coder]

According to u::
Weight3: Now you have only two coins. Weight this and found the defected coin.

Now u hve 1 coin each on two sides of balance.
How will u know which is defected coin bcoz u dnt know the defected coin is heavier or lighter!!

[psuresh1982]

I think it is not possible to find the defected weight according to your condition "u dnt know the defected coin is heavier or lighter!!". Because we are using only a balancing weighting machine not a systematic weight machine.

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suresh

[smart_coder]

Still u hve to find!!

[psuresh1982]

hi smartcoder,

Please leave your solution for this puzzle.

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suresh

[smart_coder]

Let other people also think for the solution.

[neelim]

Hi all...
Even if i want 2 look at dis site evryday atleast.im nt finding time atal..
Nw 2 jst wanted 2 hv a glance over dis..
Smart coder..
For dis problem...i m jst making a guess..jst tell me whether its correct r nt..
So,acc 2 me..
Firstly..name all the coins from 1 to 12.
On d left side weighing -- 1 2 3 10,1 2 3 11,1 4 7 10
on the right side weighig ---4 5 6 11, 7 8 9 10 , 3 6 9 12
will have these combinations( i have taken it randomly)
lets say..if its left side heavier --- lh
if its right side heavier --- rh
if both sides same --- lrs
now we will hav combinations lik dis... Lh lh lh ---- 1 heavy
lh lh rh --- 3 heavy
lh lh lrs --- 2 heavy
lh rh lh --- 10 heavy
lh rh lrs -- 11 light
lh lrs lh -- 6 light
lh lrs rh -- 4 light
lh lrs lrs -- 5 light .. And so on.. Lik dis..we cn find out d odd coin

thanks neelima

[smart_coder]

Neelam I m unable to understand ur solution.
What do u mean by:
On d left side weighing -- 1 2 3 10,1 2 3 11,1 4 7 10
on the right side weighig ---4 5 6 11, 7 8 9 10 , 3 6 9 12

[neelim]

left side right side

weighing 1 1 2 3 10 4 5 6 11
weighing 2 1 2 3 11 7 8 9 10
weighing 3 1 4 7 10 3 6 9 12

[neelim]

left side right side

weighing 1 1 2 3 10 4 5 6 11
weighing 2 1 2 3 11 7 8 9 10
weighing 3 1 4 7 10 3 6 9 12

These are weighing i vl hav on d left n right side of the weighing machine

[smart_coder]

How u deduced dat if all the three times left side balance is heavy, the coin 1 will be defected?
i.e
if in ur terminology
lh lh lh then coin 1 is defected

how u arrive at this conclusion?

[neelim]

in d case --- lh lh lh --- 1 is heavier

as u cn c 1 in the left side of weighing machine which makes diff in al 3 times

In d case --- lh lh rh --- 3 is heavier

as u cn c in d right side of weigihng machine dat makes heavy..

[psuresh1982]

Divide the 12 gold coins into (4,4,4)

1. Check the first two 4-set of coins. If both of them equal then divide the 4 into (3,1). Check the 3-set of coin with the regular 3 gold coin. If both of them equal then the remaining one is the damaged coin.
Othewise divide the 3 into (1,1,1). Now check first two 1 gold coin. If both of them equal then the third one is damaged.
If the scales move, the odd coin is the one that moves in the same direction that the three coins under consideration moved in the previous weighing.
If it moves up, it's lighter; if it moves down, it's heavier.

2. If it's not equal, take one coin from each side and switch them. One one side only, remove the other three and set them aside for later. Replace them with three gold coin from the four left on the table (now known not to be the odd one).

If the two sides are equal, the odd coin is among the three set aside. Weigh one against another, and set the third aside.
If the sides are equal, the odd coin is the third one, and it is heavier or lighter depending on which way the scales moved in the first weighing.
If the scales move, the odd coin is the one that moved in the same direction as it did in the first weighing, and it is heavier or lighter depending on whether it went down or up.

If the two sides move in different directions as in the first weighing, the odd coin is one of the two that switched places. Weigh one of the two against any of the other ten.
If both sides are equal, the odd coin is the one left out. It's heavier or lighter depending on which way the scales moved in the second weighing.
If the scales move, the coin on the scales that's under consideration is the odd one, and it is heavier or lighter depending on whether it went down or up.

If the two sides move in the same direction as in the first weighing, the odd coin is one of the three that hadn't moved from its side. Weigh one of the three against another, and set the third aside.
If the sides are equal, the odd coin is the third one, and it is heavier or lighter depending on which way the scales moved in the previous weighings.
If the scales move, the odd coin is the one that moved in the same direction as it did in the previous weighings, and it is heavier or lighter depending on whether it went down or up.


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suresh

[smart_coder]

2. If it's not equal, take one coin from each side and switch them. One one side only, remove the other three and set them aside for later. Replace them with three gold coin from the four left on the table (now known not to be the odd one).

I am unable to understand this statement.

[psuresh1982]

hi smartcoder,

It means divide the 4,4 coins into (3a,1a) and (3b,1b). Next change 1a to 1b and 1b to 1a. Also took 3a coin and replace with undamaged 3.(This is what we found in first weight).

I think you understand the next steps. If not then ask me.. I will explain.

-------------------
suresh

[sricreeper]

Possibility A :
==========
Considering that weight of one coin is heavier than the other 11 coins, we can do the following:
(A) Divide the 12 coins into 3 equal sets A,B & C (4 coins each)
(B) Weigh any two sets - Say, A & B. (Balance used 1st time)

1. Possiblity that both sets are equal, then the 3rd set has the defected coin. So you can ignore the first 2 sets.
2. Now, take the 3rd set and put two coins on either side of the balance. The one which is more in weight will come down. So, u can ignore the lighter side of the coins. (Balance used 2nd time). The two coins on the lighter side can be ignored as they are of same weight.
3. Weigh the remaining two coins, with one on either side of the balance and the one with more wight will go down. This coin is the one which is defective. (Balance used 3rd time).

(C) Suppose, sets A & B don't match, even then the set which is more in weight has the defective coin, meaning to say that either Sets-A & C or Sets-B & C can be considered as equal and hence can be ignored. Now follow the same steps (2) and (3) as in Step-B. You can find the defective coin.


Possibility B :
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Considering that weight of one coin is lighter than the other 11 coins, we need to do the same process but should take the lighter side & compare.

Hope I have given the right solution

Cheers,
Srilatha.K