Well, here's some logic then. If there are four numbers, there are only six different pairs that we can compare to look at the difference. (I.e. for A, B, C, D, theres |A-B|, |A-C|, |A-D|, |B-C|, |B-D|, |C-D|. So if we're looking at a list of six different differences between those four numbers, we are looking at a

*complete* list. So the largest difference among these numbers is 6 - this means that the largest number is 6 more than the smallest number, and two others are in between. Where are they? Well, there's also a difference of 5 somewhere, and the only way to fit this into the range is to place a third number 5 from one end and 1 from the other end. If the smallest number is x, we have two possibilities:

x

x + 1

x + 6

or

x

x + 5

x + 6

For brevity I'm just going to talk about the first one - remember that this same logic could also be applied to the second one if needed.

Now where can we fit the one remaining number? A difference of 1 is already taken, so we can't use x+2 or x+5. If we use x+3, that would give us differences of 2, 3, and 3 - no good. Using x+4, we get differences of 3, 4, and 2 - which fills in the remaining difference reqirements nicely. Great. So, we have

x

x + 1

x + 4

x + 6

What's x? Well, the sum needs to be 27. This gives us 4x + 11 = 27, yielding x = 4. This gives results 4, 5, 8, 10 as CK posted.

What about the other possibility I skipped? Repeatiing the equivalent logic for that gives us four numbers:

x

x + 2

x + 5

x + 6

Summing these gives 4x + 13 = 27 and x = 3.5. So we could say there's a solution for 3.5, 5.5, 8.5, 9.5. However most people don't use decimals when reporting their age, so we should probably reject this one.

Alternately if they do use decimals and the values are exact, this implies that all 4 girls have the same birthday, which seems unlikely but is possible in Puzzleland I suppose.

Note also that (a) there is actually some "hit and trial" lurking under the covers here; oh well; and (b) this is one of those problems that is about ten times more work to explain than it was to solve. I mean really, all I did was sketch a little number line, and most of it fell into place.

[ June 19, 2003: Message edited by: Jim Yingst ]