Positive integers $a$ and $b$ satisfy $a b=2010$. If $a>b$, the smallest possible value of $a-b$ is
(A) 37
(B) 119
(C) 191
(D) 1
(E) 397
Note that $2010=10(201)=2(5)(3)(67)$ and that 67 is prime.

Therefore, the positive divisors of 2010 are 1, 2, 3, 5, 6, 10, 15, 30, 67, 134, 201, 335, 402, 670, $1005,2010$.

Thus, the possible pairs $(a, b)$ with $a b=2010$ and $a>b$ are $(2010,1),(1005,2),(670,3)$, $(402,5),(335,6),(201,10),(134,15),(67,30)$.

Of these pairs, the one with the smallest possible value of $a-b$ is $(a, b)=(67,30)$, which gives $a-b=37$.

ANSWER: (A)