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The odds stay unchanged (Posted on 2004-06-17) Difficulty: 4 of 5
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?

See The Solution Submitted by Ady TZIDON    
Rating: 4.2727 (11 votes)

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Solution Puzzle Solution Comment 21 of 21 |
(In reply to Answer by K Sengupta)

Let m be the number of members in one of the clubs(say), and the number of members in the other club is n(say). Without loss of generality, let us assume that m>=n.

The total number of distinct pairs of members in the club having m members is comb(m, 2), while the number of distinct pairs of members in the club having n members is comb(n, 2). Also,
the total number of distinct pairs of members considering one member from the m-member club, and another from the n-member club is m*n.

Then, by the given conditions, we must have:

m*n = comb(m,2) + comb(n,2)
or, m^2 - m + n^2 - n = 2*m*n
or, (m-n)^2 = m+n .......(*)

Accordingly, the total membership (m+n) of the two clubs, which is equal to the total population of the island, must therefore correspond to a perfect square.

Let, the respective initial population of the island and the population after the increase be t^2 and (t+s)^2.

Then, we have:

(i) m(x) + n(x) = x^2, for x = t, t+s
(ii) m(x) - n(x) = x, for x = t, t+s
(iii) (t+s)^2 - t^2 = 100

Solving (i) and (ii), we have:

m(t) = (t^2 + t)/2, m(t+s) = ((t+s)^2 + t+s)/2,
n(t) = (t^2 - t)/2, m(t+s) = ((t+s)^2 - t-s)/2

Now,    
(t+s)^2 - t^2 = 100
or, s(2t+s) = 100
or, t = 50/s - s/2

Thus, s is an even integer dividing 50. For s> 10, we would have negative t. This is a contradiction. Thus, s<=10, and the only valid values of s are 2 and 10.

If s=2, then t=24, so that:
t+s = 26, and:

m(26) = 351, n(26) = 325, m(24) = 300, n(24) = 276

If s=10, then t=0, so that:

m(10) = 55, n(10) = 45, m(0) = n(0) = 0

Consequently, we have two possible solutions to the given problem:

(I) The island is initially unpopulated. After the induction of 100 persons into the island, the respective membership of the two clubs is 55 and 45.

(II) The island had a initial population of 576, with the respective membership of the two clubs being 300 and 276. After an increase in the population by 100, the island now has a population of 676, with the respective membership of the twoclubs being 351 and 325.

Edited on May 27, 2008, 4:10 pm
  Posted by K Sengupta on 2008-05-27 15:50:58

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