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

Math, Truth, and the Glory of God

This quarter in school I am talking a wide range of classes. It's nice to exercise various parts of my brain. Now more than ever I am realizing how wonderful math is. I and thankful that this is one class where almost everything is true. Unlike literature, or biology where un-biblical worldviews must be filtered with much care, math presents principals that are true and right. Finally, a class where I am not being taught against God. However it is important to note that math can be studied in a way that doesn't give glory to God. We must be careful to see the glory of God through math and not use math(or any other subject) to exalt ourselves.

"Whatever you do in word or deed, do all in the name of the Lord Jesus, giving thanks through Him to God the Father."
     -Colossians 3:17

May God give us the grace to glorify him in our education.

-Roman