Suppose you had 8 billiard balls, and one of them was slightly heavier, but the only way to tell was by putting it on a scale against another. What's the fewest number of times you'd have to use the scale to find the heavier ball?

Put 4 balls each in Pan A & Pan B find the pan having the heaviest ball. Let it be pan b. So all 4 balls in Pan A are good balls, keep it seperate. Now group the balls in pan b into two sets 2 in each pan. Find the Pan having the heaviest ball Let it be Pan A. Now you have identified 6 good balls.

Take one ball from Pan A and weigh it against any one of the 6 good balls if both are equal then the other ball is the culprit else this one should be the culprit.

Put three on each side of the scale. If they match weigh the other two. If they don't match, weigh two of the heavy three.

A good question is never answered. It is not a bolt to be tightened into place but a seed to be planted and to bear more seed toward the hope of greening the landscape of the idea. John Ciardi