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2006 Cyprus MO/Lyceum/Problem 24

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Problem

The number of divisors of the number $2006$ is

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

Solution

$2006=2\cdot17\cdot59$ . 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 $2006$ .

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_Cyprus_MO/Lyceum/Problem_24"

Category: Introductory Algebra Problems

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