the problem of the 12 stones, revisited

There are 12 stones, 11 of equal weight, 1 of unequal weight (let’s say, lighter).

By using a two-pan balance or a a pair of scales balance, can the lighter stone be found with just two measures?

Well, the correct mathematical answer is no… The maximum number of stones amid of which the lighter ball can be found with just two measures is exactly nine. In order to resolve the problem, we’d need another measure, so with three measures it is solvable.

But, even if mathematics show us a closed alley… it doesn’t mean that we can’t use this problem as a good case for using lateral thinking. This way a solution for the problem could be found… hopefully without exerting an extreme force against the rules of logic, physics… or even against the entire Universe.

So, I have bent the logic under the wind of the statement of this problem… remember 12 stones, one lighter, and just 2 measures allowed with the two-pan balance.


First of all, maybe luck is on our side, so let’s try find a solution with more than 50% odds…

One possible solution is to divide the 12 stones in three groups of four stones each one. The scheme of probabilities is:

scheme of probabilities solution to the twelve stones problem

Well, just one important note: if a measure shows that those stones weigh the same, we can then pick one of them and substitute that one for one of the remaining group (in which we know that the lighter stone must be). In the scheme I have substituted stone #12 with stone #8. This way, we can obtain one more case in which the weighting let us choose the lighter stone.

So, we end up with a 3 OK over four attempts (the scheme should have 2 or three branches, but this does not change chances), or a 3/4 = 75 % of probability of finding the lighter stone.

With the same substitution trick, we can obtain the next probabilities, beginning by weighting two equal groups of:

3 stones => 4/7= 57%

4 stones => 3/4= 75%

5 stones => 2/3= 66%

Any of these grouping strategies plus the substitution trick give us more than 50% of chances of finding the lighter stone.

Other groupings aren’t so promising:

2 stones => 1/9= 11%

6 stones => 1/5= 20%

Note that without yhe substitution trick, the 4 stones grouping is the best strategy as it gives a 50% of chances. But we were more ambitious!


My favourite solution is not in fact a solution, as it contains a fallacy.

It starts demonstrating than one can use a pan balance to weight things… without gravity: we’ll do this by showing a mathematic-physical equivalence between mass and velocity. In the usual way of using a two-pan balance, we’d obtain only an indication of whether one mass is heavier than the other, or if they have the same mass. Then we eliminate gravity from the premises (it is not even implied in the problem’s statement!): Now you see why I speak about “mass” nad not about “weight”. We can also eliminate air as to not have to count on friction forces which are very difficult to cope with. In this scenario, we can throw one mass against one of the pans at a predefined velocity -yes, we must be able to measure velocity… but anyway with gravity we have to be able to compare the level of the pans, so let’s change one capacity for the other- so when it hits the pan, it’ll transmit its force to the other pan, which in turn will make the other mass acquire a velocity… the relation between this second mass’ velocity and the former is the inverse of the relation between the masses of the objects:

relation of velocities on a zero g ambient for a two pan balance relation of velocities on a zero g ambient for a two pan balance

if m(1) > m(2) => v(1) > v(2), because: m(1)·v(1)=p(1) must be equal to p(2)=m(2)·v(2)

(September 2013 edit and digression: lineal moment is ALWAYS conserved, independently of if the collision is completely elastic or not: the latter just states that the kinetic energy is NOT always conserved… is just the “total” energy of the system which cannot vary. See this beautiful exemplification. In fact, lineal moment conservation is probably the most simple, beautiful and unbeknown law of nature).

that is:

p(1)=p(2) => m(1)·v(1)=m(2)·v(2) | if m(1)>m(2) => v(1)<v(2) & if m(1)=m(2) => v(1)=v(2)

relation between pan separation on g gravity and velocity in 0 g for a two pan balance

With this equivalence demonstrated, take the 12 stones to an environment without gravity, say the International Space Station (ISS) as I think microgravity is permissible (though I’m not sure).

Put all the stones on one of the balance’s pans, and apply a constant force to the other pan, so the stones will end flying upwards (well, you understand) – and the lighter stone would occasionally end upfront of the others. As it has less mass, and as all the stones has received the same force, it must have acquired a greater velocity:

F = m · a; if [ m(1)<m(i) | i<>1 ] & F(1)=F(i) => a(1)>a(i) | i<>1

Everything would be ok… the fallacy resides in the fact that the acceleration applied to all stones is the same: as they travel the same little distance (the distance and time they are in contact with the pan) and are thrust the same amount of time (t) (when one pan is pressed the other one is pressing the stones, and none of them stops until the other one does so), when they abandoned the pan they’ve received the same acceleration, so they’ve ended up with the same velocity… The force applied to the lighter stone has been indeed different (less) than the force applied to the other eleven stones.

What a pity!

But this is not very surprising, as by the previous demonstration of equivalence between the both methods of using a two-pan balance, we know that in fact we were just… that’s correct, weighting “all” the 12 stones, not only some of them.


We can take one of the pans, which will hopefully have a curved surface, and throw all the 12 stones to the bottom. Then we’ll produce a circular movement to the pan, until the lighter stone be expelled by the centrifugal forces. Then we can use the balance just to check the solution. Twice :)


This time we can suppose that we can use the Archimedes principle and use some recipient with water in order to measure with the balance (weigh) the displaced water of each of the stones… We do not measure the weight of the stones themselves so we hope this evict somehow the two weightings limitation.

Nonetheless, with this measures we will only know the volume of the stones… Nothing in the statement allow us to suppose that they are even of the same material, so they could for example weight all the same, but have each one different volumes.


Well, it’s time to use the Many-worlds interpretation (MWI) of the quantum mechanics. In lay terms, the hypothesis states there is a very large —perhaps infinite— number of universes, and everything that could possibly have happened in our past, but did not, has occurred in the past of some other universe or universes.

So, let’s weigh just two random choose stones… There’s at least one alternative universe in which we’d have chosen the lighter stone. So, we just have to select and observe (somehow) that particular universe in order to know the answer. We can then make a second weigh just to test the result :)

How to select and observe that alternative universe is left as an easy exercise. Clue: buy a bunch of red gummy worms.


Now, as you can imagine, it’s time to use a blackhole. And in particular we will measure the Hawking radiation of one of them, we can choose anyone – it is not important… well may be the blackhole’s mass is important, but I’m not sure. Anyway, as every schoolboy knows, the electromagnetic radiation emitted by a blackhole by effect of the Hawking radiation is as if it were emitted by a black body with a temperature that is inversely proportional to the black hole’s mass. Well, that said, let’s throw each of the stones by turn, and after each thrown, measure the radiation emitted by the blackhole. The lighter stone will produce a different footprint, so we’ll know that it is (well, was, if it’s not the last one) the lighter one. As the statement didn’t say anything about the stone/s disappearing from the accesible universe, it seems a plausible solution. Then we can use the two-pan balance to make two weights… just to complete the solution, you know. After which we can also throw the balance to the blackhole… and make this problem completely disappear from our Universe.


In this occasion, we need to situate ourselves on the synchronous orbit of a gravitational body. In this peculiar position, we’ll put six of the stones on one pan of the balance and we’ll exert a force on the other pan so as to throw the stones with enough speed to maintain them on a clockwise synchronous orbit. That counts as one measure as demonstrated on “1“, even if we don’t care for the result… Then we rotate the balance so as to point the pans to the the same orbit but counter-clockwise. Then, we repeat the operation with the other six stones, taking care that they be exactly aligned with other six, so they’ll end up crashing two by two. As they all have the same velocity, and the lighter one has less mass, it’ll be thrown away at greater velocity than the others.

September 2013 edit: again, as in method “1“, it is not important if the collision is elastic (and btw, not destroying) or not. As the lineal momentum is conserved, and both stones has same speed (as demonstrated on method “1”), the lighter stone will reverse its trajectory and the heavier will continue with the same trajectory… both at less speed.


Speaking of “6“… We do not need a gravitational body. One could become fussy at this point and say that it is not implied in the statement (even if without it the term “weight” has no sense), so we can just suppose that our Universe is finite and unlimited. It doesn’t seem so, but do not let this stop us: we will repeat the “6” procedure, but far away from gravitational bodies, and assuring that stones will not find any obstacle on their journey… as we will send six of them to one direction, and the other six to exactly the opposite direction. We’ll wait there, or at the appropriate spot of the Universe depending on its geometry, during an eternity for them to rejoin (or we’ll wait in vain, as most waits end) and collide, so as to obtain our answer. With this method we’ll also take some important cosmological clues about the shape and size of the Universe… or not. That’s the problem sometimes with eternity: it doesn’t help at all.

Hope I haven’t commit any mistake on any of the seven solutions.



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