April wrote: How many pairs $ (m,n)$ of positive integers with $ m < n$ fulfill the equation $ \frac {3}{2008} = \frac 1m + \frac 1n$? It's equivalent to: $ 9mn = 3*2008(m+n) \iff (3m-2008)(3n-2008) = 2008^2$. Thus $ 3m-2008 \equiv 3n-2008 \equiv -1 \bmod 3$. So if we get numbers $ ab = 2008$ such that $ 3m-2008 = a$ and $ 3n-2008 = b$ we want $ a \equiv b \equiv -1 \bmod$. Which happens iff $ a \equiv -1 \bmod 3$. There are $ 10$ positive numbers $ a$ such that $ a \equiv -1 \bmod 3$. Because of symmetry $ 5$ of them will have $ a > b$ so the answer is $ \boxed{5}$

We can say that some $d$ is the greatest common divisor of $m$ and $n$, so we can write $m=dx$ and $n=dy$. We will now put it in the equation: $\frac {3}{2008} = \frac {1}{dx}+ \frac {1}{dy}$ or $3dxy=2008(x+y)$. Now, $2008=2 \cdot 2 \cdot 2 \cdot 251$, so we have few cases that can satisfy this equation, because $2008$ is not divisible by $3$, that means that $x+y$ has to be divisible, so these are pairs of $x,y$ that will satisfy this condition: $(x,y)=(1,2);(1,8);(1,251);(1,1004);(4,251)$ We can find $d= \frac {2008(x+y)}{3xy}$, so $d=(1004);(753);(672);(670);(170)$ Finally $(m,n)=(1004,2008);(753,6024);(672,672 \cdot 251);(670,670 \cdot 1004);(680,170 \cdot 251)$ So, there are $5$ solutions for this equation