# Word problem on divisibility

I came across the following word problem related to divisibility:

In a certain town 2/3 of the adult men are married to 3/5 of the adult women. Assume that all marriages are monogamous (no one is married to more than one person). Also assume that there are at least 100 adult men in the town. What is the least possible number of adult men in the town? of adult women in the town?

If the number of adult men be x, and that of adult women by y, then we must have $\frac{2x}{3} = \frac{3y}{5}$. Also, the number of married couples must be divisible by 15. So I started my search with $x=45$ (just to be safe), and went up adding $15$ each time. However, even after $330$ I seemed nowhere close to an answer and decided to write a small C program to do the hard work:

#include <stdio.h>

int main()
{
int n = 45;

while (1)
{
printf("%d %d\n", 2*n/3, 3*n/5);

if( (2*n/3) == (3*n/5))
break;

n+=15;
}
return 0;
}


To my horror, the program went into a very long loop, and even after running into 10-digit numbers, there was no success. My question is: Is there really no answer to this? If not, could I have told so looking at the given conditions. And on the contrary, if there is, how can I find it?

Also, the number of married couples must be divisible by $15$.

No, it need not. Two thirds of the adult men are married, and that implies that the number of married couples is divisible by $2$, not necessarily by $3$. Three fifths of the adult women are married, and that implies the number of married couples is divisible by $3$, so altogether the number of married couples is divisible by $6$, not necessarily by $15$. From $\frac{2x}{3} = \frac{3y}{5}$, we obtain $x = \frac{9}{10}y$, so we must have $x$ divisible by $9$ and $y$ divisible by $10$. The smallest multiple of $9$ that is not less than $100$ is $108$, which leads to $120$ adult women and $72$ married couples.

• Very nice answer! I must say the first three sentences were a revelation. :-) Commented Dec 5, 2013 at 10:15
• I think $\gcd(9,10)=1$ plays a role here too (?). The statements that $x$ is divisible by $9$ and $y$ is divisible by $10$ come from there. That is $9 | 10x$ but since $\gcd(9,10)=1$, it implies that $9 | x$. Similarly for $y | 10$ as well. Am I right? Commented Oct 29, 2017 at 6:11
• @Mathmore Yes. If we have $\frac{a}{b}$ of the adult men married to $\frac{c}{d}$ of the adult women, we get $x = \frac{bc}{ad}y$, and we can only deduce $(bc) \mid x$ and $(ad)\mid y$ if $\gcd(ad,bc) = 1$. Otherwise, we must reduce the fraction, say giving $\frac{p}{q}$, then we find $p\mid x$ and $q\mid y$. Commented Oct 29, 2017 at 18:48
• @DanielFischer Nice! ${}{}{}$ Commented Oct 30, 2017 at 2:30

HINT:

As lcm$(3,5)=15$ let the number of men & women be $15x,15y$

So, the $10x$ men are married to $9y$ women

So, $10x=9y\iff \frac x9=\frac y{10}=c\implies x=9c,y=10c$

So, the number of men are $15x=15\cdot9c=135c$ we need this $>100$

No of Men : $108$

No of Women : $120$

Method: trial and error, decide upon the number of women, calculate $3/5$ of that number. when you multiply that number by $3/2$ you get the corresponding number of guys.

If you want to find $N$ number of such cases, do it by putting this as the function and if clause for the number of men variable to be integer.