You have a scale that is remarkably sturdy, and there are seven spheres of unequal mass.

No sphere has the same mass as any other.

There is (almost) no limit to the mass of any sphere, but the most massive sphere is the planet that you are standing on.

You know the exact mass of the smallest two of these spheres. One is one kilogram and one is three kilograms.

Your scale is able to fit more than one sphere on each end, however the largest sphere, the final sphere, is far too large to fit on either end of the scale.

All of the remaining spheres fall between the three-kilogram and the planet, and none are large enough to cause the planet's center of gravity to be outside its core.

We will assume that the planet-sphere does not become indented at the surface due to the mass of the other spheres, or the scale.

Your scale will register this information:

a) whether the two weighings are equal

b) if the weighings are not equal, which side weighs more, and...

c) if one side weighs more, the amount by which the weight exceeds, if within 10 percent.

Do you have enough information to determine the mass of the remaining spheres using only your scale and the weighings? If not, what about all of them except the planet?

We can safely assume that an accurate measurement would be accurate to within 0.1 kilograms, and for the sake of simplicity we won't take into account cosmic dust buildup during measurement or anything of the sort.

You are free to assume any upper bound necessary for the size of "a planet." But no shenanigans regarding planets so big they collapse into black holes or any other craziness, please.


1 Answer 1


If the masses of the six smaller spheres are (in kg) 1, 3, 5, 11, 23, 49, then there is no way to learn anything from this scale except for sorting the spheres by weight.

There is no combination of those numbers that is within 10 percent of any other combination.

  • $\begingroup$ Out of curiosity, is there a magical % value where you can't create such a list of numbers? $\endgroup$
    – Kingrames
    Sep 18, 2015 at 17:21
  • $\begingroup$ @Kingrames Only if you limit the weight of the heaviest sphere. $\endgroup$
    – Zandar
    Sep 18, 2015 at 17:22
  • $\begingroup$ hmm, I might try that with a sequel to this problem - using the mass of the planet and the rule where the center of gravity can't be outside its inner core... $\endgroup$
    – Kingrames
    Sep 18, 2015 at 17:23
  • 2
    $\begingroup$ "there is no way to learn anything from this scale except for sorting the spheres by weight" - you can also establish lower bounds on the weights. $\endgroup$ Sep 19, 2015 at 1:54

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