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

(solution)
(The explanation of the formula was a little vague, so I just posted a good solution.)
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==Solution==
 
==Solution==
<math>2006 = 2 \cdot 17 \cdot 59</math>. A number has <math>(m_1 + 2)(m_2 + 1) \cdots (m_n + 1)</math> divisors, where <math>N = p_1^{m_1} \cdots p_n^{m_n}</math>, with prime <math>p_i</math>. Thus <math>2006</math> has <math>(1+1)^3 = 8\ \mathrm{(C)}</math> divisors.
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<math>2006 = 2 \cdot 17 \cdot 59</math>. A divisor of <math>2006</math> is therefore in the form <math>2^m\cdot 17^n\cdot 59^p</math>, where <math>m\leq 1</math>, <math>n\leq 1</math>, and <math>p\leq 1</math>. There are 2 choices for <math>m</math>, 2 choices for <math>n</math>, and two choices for <math>p</math>, therefore there are <math>2\cdot 2\cdot 2=\boxed{8}</math> divisors of <math>2006</math>.
  
 
==See also==
 
==See also==

Revision as of 15:17, 2 March 2008

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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