COINS TO BE DIVIDED INTO THREE GROUPS.

1ST ITERATION (27/3=9 COINS); 2ND ITERATION (9/3=3 COINS); 3RD ITERATION (3 INDIVIDUAL COINS)

MINIMUM MOVES = 3 (IF UR LUCKY)

1)make 3 groups of 9, if on 1st try u weigh equal weights, then 3rd group has odd coin.

2)make 3 groups of 3 coins from the 3rd group i.e from 9 coins which u separated after 1st iteration.

if u lucky and weigh equal weights again, then 3rd group of 3 coins has the odd coin.

3)if u r again lucky enuf to weigh equal coins, then the coin left out is the odd coin.

MAXIMUM MOVES = 4 (EVEN UR NOT SO LUCKY)

1)make 3 groups of 9, if d weights are unequal, then replace one pan with 3rd group,

so it takes 2 TRIES, if in 1st try left pan was heavier and after 2nd try both become equal then u come to know that the group of coins in left pan has the odd coin OR if d left pan is still heavier u obviosly know that is d one having odd coin and also very importantly we can get to know if .... the coin is of 11 g (if pan is heavier)... else of 9g (if pan is lighter) when compared with the other pan.

2)make 3 groups of 3 coins from the 3rd group i.e from 9 coins which u separated after 1st iteration.

now u very clearly know whether the coin is light or heavy ..... so u need not go for 2 tries.

in just one try. if u weigh equal, then 3rd group else which ever is (heavier or lighter) as u already know the weight of odd coin.

so just 1 TRY

3) again just 1 try needed, weigh any 2 coins and can clearly get d odd coin in just 1st try.

THUS ONLY 4 MOVES ARE ENOUGH, IF UR NOT LUCKY ENOUGH TO GET IT IN 3 MOVES.