20) How many ordered pairs of integers $(a, b)$, with $1<a<2000,1<b<2000$ are there such that the least common multiple of $a$ and $b$ is equal to 2000?
(A) 14
(B) 20
(C) 24
(D) 40
(E) 48.
20) The answer is (C)

Since $2000=2^{4} 5^{3}$, it must be that $a=2^{x} 5^{y}, b=2^{s} 5^{t}$ where the maximum of $x$ and $s$ is 4 and the maximum of $y$ and $t$ is 3. Note that it is not possible to have both $x=4$ and $s=4$ simultaneously, since, if $y=3$ or $t=3$, one of the numbers $a$ and $b$ would have to be equal to 2000. Suppose that $x=4$. Then $y=0,1,2$, since $a \neq 2000$. It follows that $t=3$ and, since $b$ cannot be equal to 2000 either, $s=1,2$, 3. Therefore, in this case, there are $3 \times 4=12$ possibilities for the pair $(a, b)$. Symmetrically, if $s=4$, there are another 12 possibilities, for a total of 24 possibilities.