Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 24"
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==Problem== | ==Problem== | ||
| − | + | The number of divisors of the number <math>2006</math> is | |
| + | |||
| + | A. <math>3</math> | ||
| + | |||
| + | B. <math>4</math> | ||
| + | |||
| + | C. <math>8</math> | ||
| + | |||
| + | D. <math>5</math> | ||
| + | |||
| + | E. <math>6</math> | ||
==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. |
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=23|num-a=25}} | {{CYMO box|year=2006|l=Lyceum|num-b=23|num-a=25}} | ||
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| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 17:26, 15 October 2007
Problem
The number of divisors of the number 
is
A. 
B. 
C. 
D. 
E. 
Solution
. A number has 
divisors, where 
, with prime 
. Thus has 
divisors.
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 | ||
