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There are 9 coins. Out of which one is odd one i.e weight is
less or more. How many iterations of weighing are required
to find odd coin?
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Answer Posted / saint
3 Iterations are required if you do not know if the odd
coin is lighter or heavier when you start. This is a nice
simple to follow process;
Divide coins into three piles of three, A, B & C
Weighing 1
-----------
Weigh piles A and B, and note which is the heaviest, or if
equal weight
Weighing 2
-----------
Weigh piles A and C, and note which is the heaviest or if
equal weight
If A does not equal both B and C, then A contains the odd
coin.
If A = C but does not equal B then B contains the odd
coin.
If A = B but does not equal C then C contains the odd
coin.
Having established which pile contains the odd coin, then
determine if it the odd pile is lighter or heavier by
looking at the results of the first two weighings
A = Odd pile
--------------
If A is the odd pile, then if it weighed less than B and C
in the first two weighings, then the odd coin is lighter,
else it is heavier
B = Odd Pile
------------------
Use the results of Weighing 1 and compare B with A. If A
was the lighter pile, then the odd coin is Heavier than the
rest, if A was heavier then the odd coin is Lighter
C = Odd Pile
------------------
Use the results of Weighing 2 and compare C with A. If A
was the lighter pile, then the odd coin is Heavier than the
rest, if A was heavier then the odd coin is Lighter
Weighing 3
-----------
Now take the pile that contains the odd coin, take two
coins out and weight them against one another
If the coins match, then the remaining coin is the odd coin
Else if the coins do not match, then if we are looking for
a lighter coin, then the lighter coin is odd, and if we are
looking for a heavier coin, then the heavier coin is odd!
-------------------------------------
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