BrainTeaser 1

[Wrahn]

1. Take the 24 and divide it up into 3 groups of 8 weigh two of the groups. If one group is heavier than that group has the ball in it, if they are even the third group of eight has the ball in it

2. Take the group of eight with the ball in it and add a ball from one of the eliminated balls (from the previous run), divide into 3 groups of three and weigh 2 against each other. If they weigh the same the ball is in the third group, if not then the heavier has the ball

3. Take the group with the ball in it and weigh two balls against each other. If the balls weigh the same then the ball is the third ball. The heavier is the ball if they don’t.

Hey Thallas, I think they make some INT boosting gear that even bards can wear ">

[Herbert Dypp]

There are 24 identical balls, but one is heavier than all the others.

You have a set of balance scales.

With a maximum of three weighings separate the heavier ball from the rest.

[Thallas]

Tough one, I can do it in 4 easy enough, I'm sure there's a trick to it, but bards dont have much int

[Herbert Dypp]

a little more complex than necessary Wrahn, but it still solves the problem

[Jaffod]

playing with your balls again?

[Riverdane]

weigh 9 vs 9 if they balance, discard all 18, keeping 6
if one is heavier, keep the heavier 9 and discard the rest.


If One was heavier:

take your 9 and weigh 3 vs 3, if they balance, discard all 6, if they dont, take the heavier three.

You now have 3, weigh one vs one. if they balance, you didn't weigh the heavy one. if they dont, you have the heavy one on the scale.


If the 9 vs 9 balanced:

you have 6 left over, weigh two vs two.
If they balance, discard, keeping two, easy to then weigh one vs the other.

If they dont balance, you know which is the heavier two, weigh one vs the other, same result

[Herbert Dypp]

okay we have 2 out of the three ways I know of doing this (of course there may be more)...