Difference between revisions of "1993 AHSME Problems/Problem 15"
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== Solution == | == Solution == | ||
| − | <math>\fbox{D}</math> | + | Start with the facts that all polygons have they're exterior angles sum to 360 and the exterior and interior angles make a linear pair of angles. So our goal is to find the number of divisors of 360 to make both the interior and exterior angles integers. The prime factorization of 360 is <math>2^3 * 3^2 * 5</math>. That means the number of divisors is 4*3*2 = 24. But we're not done yet. We cannot have a 1 or 2 sided polygon so we subtract off two bringing us to our final answer of 22 <math>\fbox{D}</math>. |
== See also == | == See also == | ||
Revision as of 21:42, 8 January 2015
Problem
For how many values of 
will an -sided regular polygon have interior angles with integral measures?
Solution
Start with the facts that all polygons have they're exterior angles sum to 360 and the exterior and interior angles make a linear pair of angles. So our goal is to find the number of divisors of 360 to make both the interior and exterior angles integers. The prime factorization of 360 is 
. That means the number of divisors is 4*3*2 = 24. But we're not done yet. We cannot have a 1 or 2 sided polygon so we subtract off two bringing us to our final answer of 22 
.
See also
| 1993 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
