2006 Cyprus MO/Lyceum/Problem 24

Revision as of 16:17, 2 March 2008 by 1=2 (talk | contribs) (linking Divisor to the article.)

Problem

The number of divisors of the number $2006$ is

A. $3$

B. $4$

C. $8$

D. $5$

E. $6$

Solution

$2006 = 2 \cdot 17 \cdot 59$. 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 two choices for $p$, therefore there are $2\cdot 2\cdot 2=\boxed{8}$ divisors of $2006$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 23
Followed by
Problem 25
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