Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 24"
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(The explanation of the formula was a little vague, so I just posted a good solution.) |
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==Solution== | ==Solution== | ||
− | <math>2006 = 2 \cdot 17 \cdot 59</math>. A | + | <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== | ==See also== |
Revision as of 15:17, 2 March 2008
Problem
The number of divisors of the number is
A.
B.
C.
D.
E.
Solution
. A divisor of is therefore in the form , where , , and . There are 2 choices for , 2 choices for , and two 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 |