7. How many pairs of positive integers $a$ and $b$ with $a<b$ are there such that

$$
\frac{1}{a}+\frac{1}{b}=\frac{4}{15} ?
$$

A) 1
B) 2
C) 3
D) 4
E) 5
A7. C) 3 Because $a\frac{1}{b}$. In particular, $\frac{1}{a}$ must be more than half of $\frac{4}{15}$: $\frac{1}{a}>\frac{2}{15}$. From this, it follows that $a<\frac{15}{2}<8$. Moreover, $a$ must be at least 4, otherwise $\frac{1}{a}$ would be greater than $\frac{4}{15}$. So we have four possibilities for $a$: 4, 5, 6, and 7. For $a=4$, we find $\frac{1}{b}=\frac{4}{15}-\frac{1}{4}=\frac{1}{60}$ and thus $b=60$. For $a=5$, we find $\frac{1}{b}=\frac{4}{15}-\frac{1}{5}=\frac{1}{15}$ and thus $b=15$. For $a=6$, we find $\frac{1}{b}=\frac{4}{15}-\frac{1}{6}=\frac{1}{10}$ and thus $b=10$. For $a=7$, we find $\frac{1}{b}=\frac{4}{15}-\frac{1}{7}=\frac{13}{105}$ and in this case the solution $b=\frac{105}{13}$ is not an integer. In total, we thus find 3 solutions.