# Red, White, Blue coin weighing puzzle

Infront of you are six coins, and a sensitive balance beam.

Two of the six coins are marked with red dots. Two with white. Two with blue.

In each pair of coloured coins, one of the coins is slightly heavier (meaning there are three heavy coins, and three light coins). All the heavy coins have identical weights, and all the light coins have identical weights too. Your job is to separate them into two piles: All the heavy coins in one pile, all the light coins in another, (and to do this in the minimum number of weighings).

Other than the slight weight variance, and the dot markings, the coins are visually identical; there are no external clues as to which the heavier of each pair is. The weight discrepancy is so slight that only the balance beam is sensitive enough to differentiate.

What is the minimum number of weighings to separate the coins into two sets?

## Three weighings

It should be pretty obvious that we can do it with three weighings. We chose a colour and place one coin on each side of the balance. The heavier coin of each pair will go down. We do this three times. Simple. Can we do better?

## Two weighings

It’s possible to perform the task with just two weighings, but there is a caveat for one scenario (but, if you read the problem definition carefully, it does not violate the requirements: Our task is simply to correctly

*separate*the coins into two piles of three, that’s it).To help track what’s going on we can identify the heavier coins with the upper case letters

**RWB**, and the lighter coins with the lower case letters**rwb**.Advertisement:

## Strategy

Without loss of generality, select two of the colours, and place one coin of each colour on opposite sides of the balance beam. Two things will happen: The beam will balance, or the one side will go down.

If the beam balances we’ve got one heavy coin, and one light coin, on each side (we’ll come back to this), but if one side goes down, we know

*both*coins on that side are the heavy coins (there are no other combinations). Having determined the two heavy**RW**coins, it’s a simple matter to perform a second, single,**Bb**weighing to determine which of the blue coins is heavier. Problem solved!Rw rW RW rw

What happens if our first weighing balances? (above left) In this case we take off one colour (for instance white), and replace it with the third colour (blue). Again one of two things could happen: The beam could balance, or one side could go down. If one side goes does down, we know that both coins on that side are the heaver coins, so we have determined

**RB**, and can apply this logic to the first weighing to deduce that the heavier white coin**W**has to have been on the*opposite*side to the heavier red coin on the first weighing.What happens if the beam balances the second time? In this case, we know the coins were matched opposite on each weighing.

Rw rW

Rb rB

Rb rB

If this happens, we can easily able to separate the coins into two sets: one containing the heavier coins, and one containing the lighter coins (the non-matching coin on the opposite pan in each weighing is the same mass), it’s just that

*we don’t know which set is the heavier set, and which the lighter set!*This is the caveat mentioned above. We’ve correctly solved the puzzle and separated the coins, it’s just that we don’t know which set is the heavier set (if we are unlucky enough to select the coins this way, we’d need a third weighing if we want to determine this extra information).