Scales question. What is the answer?

[T0 Th3 M4X]

There's an alternate way to solve the original problem.

You can weigh them 3vs3. If the heavier ball is in one of those two groups in will tip the scale to that side, then you work from that group of 3.  If it's not, you know it's one of the remaining 2 balls.  So you can actually get it in two if it happens to be in the set of two, but no more than 3 if it isn't.

Good thinking.

Thank you sir.

There's actually an alternate to this question.  Instead of stating that the ball is heavier, you can ask if the weight is different, but not specified as higher or lower.  You just can't use too many versions or someone will end up slapping you silly.

[Javaman]

There's an alternate way to solve the original problem.

You can weigh them 3vs3. If the heavier ball is in one of those two groups in will tip the scale to that side, then you work from that group of 3.  If it's not, you know it's one of the remaining 2 balls.  So you can actually get it in two if it happens to be in the set of two, but no more than 3 if it isn't.

Actually, you are guaranteed to get it in two steps if you start by weighing them 3x3.

If one of the sets of three is heavier, all you need to do is pick any two balls from the heavier trio and weigh them against each other. If the scale tips for one, you obviously have identified the heavier ball. If the scale remains perfectly balanced, then the ball you didn't weigh must be the heavier one.

So, two steps.

[T0 Th3 M4X]

There's an alternate way to solve the original problem.

You can weigh them 3vs3. If the heavier ball is in one of those two groups in will tip the scale to that side, then you work from that group of 3.  If it's not, you know it's one of the remaining 2 balls.  So you can actually get it in two if it happens to be in the set of two, but no more than 3 if it isn't.

Actually, you are guaranteed to get it in two steps if you start by weighing them 3x3.

If one of the sets of three is heavier, all you need to do is pick any two balls from the heavier trio and weigh them against each other. If the scale tips for one, you obviously have identified the heavier ball. If the scale remains perfectly balanced, then the ball you didn't weigh must be the heavier one.

So, two steps.

Well played sir.

[Javaman]

Actually, you are guaranteed to get it in two steps if you start by weighing them 3x3.

If one of the sets of three is heavier, all you need to do is pick any two balls from the heavier trio and weigh them against each other. If the scale tips for one, you obviously have identified the heavier ball. If the scale remains perfectly balanced, then the ball you didn't weigh must be the heavier one.

So, two steps.

Well played sir.

Thank you, but I'll admit I probably wouldn't have figured it out if you hadn't suggested the 3x3 weighing.

Team effort, lol.

[T0 Th3 M4X]

Well played sir.

Thank you, but I'll admit I probably wouldn't have figured it out if you hadn't suggested the 3x3 weighing.

Team effort, lol.

Now I'm stuck in "why didn't I think of that" mode.

If someone could figure it out in one, my head would probably explode.

[Fireball]

Make a ring-shaped holder that is suspended from the middle and place the balls equally spaced around the perimeter. Some means of restraining the balls would be needed, of course. The ring will tilt. The lowest portion of the ring will be the location of the heaviest ball.

[T0 Th3 M4X]

Make a ring-shaped holder that is suspended from the middle and place the balls equally spaced around the perimeter. Some means of restraining the balls would be needed, of course. The ring will tilt. The lowest portion of the ring  will be the location of the heaviest ball.

My head my explode anyway.  The dogs are growling at each other over bones.  After awhile, my brain gets frazzled from listening to them.