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

Latest revision as of 12:15, 26 April 2008

The number of divisors of the number $2006$ is

$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6$

$2006=2\cdot17\cdot59$ . 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 2 choices for $p$ . Therefore, there are $2\cdot2\cdot2=\boxed{8}$ divisors of $2006$ .

@@ Line 1: / Line 1: @@
 ==Problem==
-{{empty}}
+The number of divisors of the number <math>2006</math> is
+<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6</math>
 ==Solution==
-{{solution}}
+<math>2006=2\cdot17\cdot59</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 2 choices for <math>p</math>. Therefore, there are <math>2\cdot2\cdot2=\boxed{8}</math> divisors of <math>2006</math>.
 ==See also==
 {{CYMO box|year=2006|l=Lyceum|num-b=23|num-a=25}}
+[[Category:Introductory Algebra Problems]]