On a certain island each of the inhabitants is a member of one of the two existing clubs.

The membership distribution is such that when two random people meet, the probability of those two belonging to the same club is equal to the probability of them belonging to distinct clubs.

When 100 newcomers arrive on the island and each enrolls in one of the two clubs, the distribution still retains this feature. How many people belong to either club?

(In reply to A trivial solution by Erik O.)

The number of members in Club Alpha and the number of members in Club Beta are always adjacent numbers in the series of trianglular numbers: 0, 1, 3, 6, 10, 15, 21,... [*]

Adjacent triangular numbers add up to be squares of integers.

The total number of people on the island are therefore always a square number.  Since we added 100 more people, we have to find pairs of triangular numbers whose sums are 100 apart.

Since there are no adjacent pairs of square numbers with a difference of 100 we have to look at the squares two apart in the sequence of square numbers:  25²-24² = 576 and 26²-25² - 676, so our total populations on the island are 576 to begin with and 676 at the end.

The two adjacent triangular numbers that add up to 576 are 276 and 300, and the two adjacent triangular numers which add up to 676 are 325 and 351.

So, the island started with a population of 576 with 276 in Club Alpha and 300 in Club Beta, then when the 100 newcomers showed up the population of the island jumped to 676 with 325 in Club Alpha and 351 in Club Beta.  (I suppose that the ending club memberships could be swapped and still have a valid answer.)

[*] Per SilverKnight's request, this comment has been updated to include why the club memberships are adjacent triangular numbers.

In order to keep the handshake probabilities even, there must be the same number of pathways between members of Club Alpha and Club Beta as there are pathways within Club Alpha and within Club Beta combined.  Mathematically speaking, A•B = A(A-1)/2 + B(B-1)/2.  Quite coincidentally (or not) the series of triangular numbers can be expressed as (x²-x)/2 or x(x-1)/2.

AB = A²-A + B²-B
          2

 

2AB = A²-A+B²-B
 
2 = A²-A + B²-B
        AB
 

2 = A-1 + B-1
     B     A

This last formula only holds true for adjacent numbers in the sequence of triangular numbers.

Edited on June 17, 2004, 5:29 pm

Edited on June 17, 2004, 5:29 pm

Posted by Erik O. on 2004-06-17 15:41:35