7. How many pairs of positive integers $(a, b)$ are there such that $a$ is a divisor of $b$ and moreover, $a + 2b = 1010$?
(A) 503
(B) 1
(C) 2
(D) 3
(E) 504
Problem 7. The correct answer is (D).

Let $b=k a$, where $k$ is a positive integer, the equation can be rewritten in the form $a(1+2 k)=1010$. Therefore, $1+2 k$ is a divisor greater than 1 of 1010, which must necessarily be odd. The number 1010 factors as $2 \cdot 5 \cdot 101$, so $1+2 k$ can be equal to 5, 101, or $5 \cdot 101=505$, corresponding to $k=2$, $k=50$, $k=252$. Thus, there are three solutions: $(a, b)=(202,404), \quad(a, b)=(10,500), \quad(a, b)=(2,504)$.