2006 Cyprus MO/Lyceum/Problem 24

Problem

The number of divisors of the number $2006$ is

$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6$

$2006=2\cdot17\cdot59$ . A divisor of $2006$ is therefore in the form $2^m\cdot 17^n\cdot 59^p$ , where $m\leq 1$ , $n\leq 1$ , and $p\leq 1$ .

There are 2 choices for $m$ , 2 choices for $n$ , and 2 choices for $p$ . Therefore, there are $2\cdot2\cdot2=\boxed{8}$ divisors of $2006$ .