# Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 24"

## 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 $2006$.