8 Marbles and a Balance Scale Puzzle



Let’s say you have 8 marbles and a two-pan balance.

All of the marbles look the same. Each marble weighs 2.0 grams except for one, which is slightly heavier at 2.05 grams.

How would you find the heaviest marble if you are only allowed to weigh the marbles 2 times using the balance scale?


Scroll down for the solution…














What if we just divide the 8 marbles into 2 groups of 4 each, and we put 4 marbles on one pan and 4 marbles on the other. After the first weighing, we’d know which group of 4 marbles is heavier, which means that we also know that the heaviest marble is in that group.

Now, we’ve narrowed it down to 4 marbles, and we know one of those marbles is the heaviest – and this is after one weighing. Can we find out the heaviest marble in one more weighing with just 4 remaining marbles?

What if we just compared 2 marbles? If one of those was the heavier marble then we would know which marble is the heaviest in 2 weighings. But, one of those 2 marbles is not necessarily the heavier one – and we don’t know which of the other 2 marbles that were not weighed is heavier.

What if we just put 2 marbles on each pan and do another weighing? One side would be heavier, and we would be able to narrow it down to 2 marbles – but we still don’t know which of those 2 marbles is heavier. This would require one more weighing.

So this solution doesn’t work… it forces us to do a third weighing, but we’re only allowed to do two…


Here we go, the real solution:

We put 3 marbles on each side of the scale for a total of 6 marbles that are being weighed. We leave 2 marbles off the scale. Then, we compare the 6 marbles… There’s two possible outcomes:

  • If one side is heavier than the other then we only have 3 marbles left. Then, in the second weighing, we can compare 2 of those 3 marbles to each other. If they are the same weight, then the 3rd is the heaviest. If one is heavier than the other, then we’ve succeeded in just two weighings.
  • If, when comparing the 6 marbles, we find that both sides are equal, then we know that the heaviest marble has to be in the 2 marbles that are not on the scale. This means that we only have to compare on those 2 remaining marbles and we have the heaviest marble in only two weighings.

Problem solved.