Monday puzzle: measuring ball bearings

That would take far more time than the three simple divisions by two. So, I don’t see how that’s easier.
Also, are you saying to separate the total into equal groups of 3 and 8? Like 20 groups of 3 and 20 groups of 8? That is assuming that the original total can be evenly divided into groups of 11.
Also a ratio of 3:8 (or 9:24) would give you a fraction of 9/33, not 9/24.
Also…regarding the original problem, are we to make the assumption that the total is divisible by 8? If it isn’t, then one of those binary divisions will not split equally. So, are we making that assumption, are just going with a rough estimate of 9ozs?

I agree with Paul & William.

I would describe the corrected version of Roberto’s approach a little differently. Divide the ball bearings into groups of 8 then take 3 from each group.

I think Roberto’s solution is definitely easier to describe. It would also be easier to implement if the number is sufficiently small. Also to its credit, it doesn’t require the balance. Comming up with this solution indicates the ability to think outside the box (here the box is the suggestion to use the balance). I must admit that I let the suggestion influence me and I only came up with the solution that was posted (no wonder lawers like “leading the witness”, and objecting to others doing so).

However, Roberto made another mistakes besides the one corrected by Paul and William. He assumes all the ball bearings are of equal size and weight, but I don’t see that as given anywhere. It is however a reasonable assumption for a box of ball bearings.

Even if all the balls were of the same size (which is really irrelevant for this problem) they could be of different materials (say stainless, brass and teflon). In this case, each type would have to be divisible by 8 for the counting method to work. The balance method might be able to work without that limitation, but could still have some problems.

If they are all the same size, to be able to get 9oz (without splitting one), the number must be divisible (I assume this is how Roberto gets his presumption of divisibility – but it only applies to 8, not 3!)by 8 so either method would work. For the smallest numbers, each would have to weigh 3oz (n=8),1.5oz (n=16), 1oz (n=24), .75oz (n=32), .6oz (n=40), .5oz (n=48)…

It seems the intent of the problem is to have ball bearings light enough that moving one back and forth on the balance won’t have a clear effect (in other words, the weight of one is below the precision of the balance or the desired precision). With that in mind, I would change the problem to one with 24oz of sand. This would avoid the problem of discreteness, but would also make the counting method inacurate (grains of sand vary in weight)as well as prohibitive.

Any liquid or powder could replace the sand, but they tend to have problems due to surface tension and electrostatics. They are generally hard to handle without losses at every step. But Mercury could work.