Cool little math problem

In a given town monogamous 2/3 of the men are married to 3/5 of the women. There are at least 100 men in the town.

What is the least possible number of men and women in the town?

m= number of men, w=number of women
(2/3)m = (3/5)w , we want lowest integer for m and w

m = (9/10)w,     with m greater-than or equal to 100
w = (10/9)m

Since w = (10/9)m,      9 divides m

Lowest number of men that is divisible by nine is 108

So we have that w = (10/9)(108) = 120

we conclude that:
(2/3)(108)= 72 men
(3/5)(120)= 72 women