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

Latest revision as of 13:15, 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 $2006$ .

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 The number of divisors of the number <math>2006</math> is
-A. <math>3</math>
+<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6</math>
-B. <math>4</math>
+==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>.
-C. <math>8</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>.
-D. <math>5</math>
-E. <math>6</math>
-==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.
 ==See also==