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 number has $(m_1 + 2)(m_2 + 1) \cdots (m_n + 1)$ divisors, where $N = p_1^{m_1} \cdots p_n^{m_n}$ , with prime $p_i$ . Thus $2006$ has $(1+1)^3 = 8\ \mathrm{(C)}$ divisors.

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 23	Followed by Problem 25
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2006 Cyprus MO/Lyceum/Problem 24

Problem

Solution

See also