posted on Aug, 2 2015 @ 07:35 AM
a reply to: Shadoefax
All right. My solution enables one to find the odd ball with three sessions... or only two if you are lucky. But it only works if the odd ball is
heavier... Anyway, here's how it goes.
Say the balls are numbered 1 to 12. We don't know which in these is the odd one out. The trick is to weight only ten of these balls - so you set aside
#11 and #12, and you put five balls on one side of the scale and five others on the other side. This constitutes your first weighting session.
-If the scale balances out, then the odd ball is not amongst the ten balls present. So it is either #11 or #12. Put these two on each side of the
scale and you'll find which is heavier, thus you solve the problem in two sessions.
-If the scale leans on one side, then the odd ball is located in the heavy-most group of five. Discard all balls but the group of five. Set one ball
aside - you end up with four balls. Divide them in two couples, set them on each sides of the scale, and weight them (second session).
---if the scale balences out, the odd ball is the one you've just set aside.
---if the scale leans on one side, the suspect ball is located in the heavy-most couple. Discard all balls except the suspect couple. Place the two
suspect balls on each sides of the scale (a third and final session) to determine which is heavier.
How's that? Am I close?