How many pairs $(m, n)$ of positive integers with $m<n$ fulfill the equation

$$
\frac{3}{2008}=\frac{1}{m}+\frac{1}{n} ?
$$

## Answer: 5.
Let $d$ be the greatest common divisor of $m$ and $n$, and let $m=d x$ and $n=d y$. Then the equation is equivalent to

$$
3 d x y=2008(x+y) \text {. }
$$

The numbers $x$ and $y$ are relatively prime and have no common divisors with $x+y$ and hence they are both divisors of 2008. Notice that $2008=8 \cdot 251$ and 251 is a prime. Then $x$ and $y$ fulfil:

1) They are both divisors of 2008.
2) Only one of them can be even.
3) The number 251 can only divide none or one of them.
4) $x<y$.

That gives the following possibilities of $(x, y)$:

$$
(1,2),(1,4),(1,8),(1,251),(1,2 \cdot 251),(1,4 \cdot 251),(1,8 \cdot 251),(2,251),(4,251),(8,251) \text {. }
$$

The number 3 does not divide 2008 and hence 3 divides $x+y$. That shortens the list down to

$$
(1,2),(1,8),(1,251),(1,4 \cdot 251),(4,251) \text {. }
$$

For every pair $(x, y)$ in the list determine the number $d=\frac{2008}{x y} \cdot \frac{x+y}{3}$. It is seen that $x y$ divides 2008 for all $(x, y)$ in the list and hence $d$ is an integer. Hence exactly 5 solutions exist to the equation.