Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 24"
(linked to a redirect...) |
I like pie (talk | contribs) m (Standardized answer choices; minor edits) |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | The number of | + | 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== | |
+ | <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== | ==See also== |
Latest revision as of 13:15, 26 April 2008
Problem
The number of divisors of the number is
Solution
. A divisor of is therefore in the form , where , , and .
There are 2 choices for , 2 choices for , and 2 choices for . Therefore, there are divisors of .
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 |