in continuation of #23546
If there are 27 balls ( let us say lead shots) of which one is heavier than the rest, we need to weigh those balls JUST THRICE to locate the heavier ball.
Tip:
We will weigh 9 balls in each side of the balance leaving out 9 balls.
If they balance, then the heavier ball is among the 9 that got left out.
Thereafter repeat the same procedure as in post #23546.
If the two sides do not balance again we will weight the balls in the heavier side as in the post #23546.
By extending this argument, if we have 81 balls on which 80 are identical in weight, we can locate the heavier one weighing them JUST FOUR TIMES.
Tip:
Place 27 balls in each side of the scale, keeping aside the remaining 27.
Repeat the same procedure as given above.
If we have 243 balls and just one ball is heavier we need to weigh them JUST FIVE times.
Tip.
Place 81 balls in each side of the scale and repeat the weighing as shown earlier.
So even if we have 729 balls with one ball heavier than the others, we can still find out that ball by weighing them JUST SIX times.
Amazing isn't it?
I would like to put the result in a capsule form!
If we have 3 to the power of n balls, then we need to weigh those balls just n times!