## 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?

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*Scroll down for the solution…*

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## 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.