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

Revision as of 16:17, 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 $2006$ .

@@ Line 13: / Line 13: @@
 ==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.
+<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==

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

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

Revision as of 16:17, 2 March 2008

Problem

Solution

See also