23. Find the number of ordered pairs ( $m, n$ ) of positive integers $m$ and $n$ such that $m+n=190$ and $m$ and $n$ are relatively prime.
23. Answer: 72
First we find the number of ordered pairs ( $m, n$ ) of positive integers $m$ and $n$ such that $m+n=190$ and $m$ and $n$ are not relatively prime.
To this end, write $m=k a$ and $n=k b$, where $k, a$ and $b$ are positive integers with $k>1$. Since $m+n=190$, we see that $k$ is a factor of $190=2 \times 5 \times 19$ with $k \neq 190$. We consider six cases:
(i) $k=2$. Then $a+b=95$, and there are 94 such pairs $(a, b)$ of $a$ and $b$ such that the equation holds.
(ii) $k=5$. Then $a+b=38$, and there are 37 such pairs ( $a, b)$ of $a$ and $b$ such that the equation holds.
(iii) $k=19$. Then $a+b=10$, and there are 9 such pairs $(a, b)$ of $a$ and $b$ such that the equation holds.
(iv) $k=10$. Then $a+b=19$, and there are 18 such pairs ( $a, b$ ) of $a$ and $b$ such that the equation holds.
(v) $k=38$. Then $a+b=5$, and there are 4 such pairs $(a, b)$ of $a$ and $b$ such that the equation holds.
(vi) $k=95$. Then $a+b=2$, and there is 1 such pair $(a, b)$ of $a$ and $b$ such that the equation holds.
It follows from the above cases that the number of ordered pairs $(m, n)$ of positive integers $m$ and $n$ such that $m+n=190$ and $m$ and $n$ are not relatively prime is $94+37+9-18-4-1=117$.

Since the total number of ordered pairs $(m, n)$ such that $m+n=190$ is 189 , we conclude that the required number of ordered pairs $(m, n)$ where $m$ and $n$ are relatively prime is 189 $-117=72$.