You have 8 Sholay coins which are all the same weight, except for one which is slightly heavier than the others (you don’t know which coin is heavier). You also have an old‐style balance, which allows you to weigh two piles of coins to see which one is heavier (or if they are of equal weight). What is the fewest number of weighings that you can make which will tell you which coin is the heavier one?

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3rd time , we’ll get the heavier coin . is it rgt ???

Good attempt Navaneetha! Please optimize the solution.

7

One possible solution. Please optimize the solution.

Twice. Put 3/3 each coin on each balance.

Condition-1: if the coins placed on the weighing pan are equal, then the heavier coin lies in the remaining two, weight the two, you’ll get the answer.

Condition-2: If one side of the pan containing 3 coins is slightly heavier, then remove the 3 coins and put 2 on the pans. If both show equal balance, the remaining coin is the heavier one, or the one heavier will show deflection during weighing.

You hit the bull eye! A neat and clean solution!

thanks a lot.!! I am looking forward, for the upcoming questions!

f the coins placed on the weighing pan are equal, then the heavier coin lies in the remaining two, weight the two, you’ll get the answer.

”it is similar 2 the 8 marble puzzle”

Good One!

in 2 steps

as explained in the 2nd comment