Problem 174 Solution

m = men before increase
w = women before increase
m' = men after increase
w' = women after increase
d = m-w
d' = m'-w'

The fact that a random drawing will result in a date 50% of the time tells us that 2*(m/(m+w))*(w/(m+w-1)) = 0.5

(m/(m+w))*(w/(m+w-1)) = .25

m/(m+w) = (m+w-1)/(4w)

4mw = (m+w)*(m+w-1) 4mw = m² + mw -m + mw + w² - w

4mw = m² + w² -2mw - m - w

(m-w)² = m + w, or d² = m+w

By the same reasoning:

(m'-w')² = m' + w', or d'² = m'+w'

We are also given:

m'+w'-m-w=100

d'² - d² = 100

(d'+d)*(d'-d) = 100

We know that d and d' must be integers so the only possible values for d'+d and d'-d are: (100,1), (50,2), (25,4), (20,5), (10,10).

Let's consider the first possibility of (100,1). If this were the solution then d'+d=100 and d'-d=1. Adding the two equations together yields 2d'=101 --> d'=101/2. However since m' and w' are integers then d' is also an integer and can not be 101/2. For the same reason (25,4) and (20,5) are not possible because the sum of the two factor is odd.

Let's consider (10,10). Then d'+d=10, d'-d = 10 --> d=0 --> m-w=0. However the problem stated m>w, so (10,10) doesn't work.

That leaves only (d'+d,d'-d)=(50,2). Adding the two equations gives us d'=m'-w'=26.

Remember, d'² = m'+w', so m'+w'=26² = 676.

Remember also that m'-w'=26

Adding the two equations give us 2m'=702 --> m'=351 --> w'=325.

So after in increase there are 325 women and 351 men. From here it can also be easily found that before the increase there were 300 men and 276 women.

Thanks to Ady Tzidon for suggesting a very similar problem and to Rob Pratt for this solution (which is better than how I initially solved it).