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Polygon

by shashank.ism » Mon Feb 08, 2010 9:13 am
Consider a regular polygon of p sides .The number of values of p for which the polygon will have angles whose values in degrees can be expressed in integers?

a) 24
b) 23
c) 22
d) 20
e) 21
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by thephoenix » Mon Feb 08, 2010 9:32 am
formula for sum of int <s of apolygon is
(N-2)*180
for regular polygon
each <s is = (N-2)*180/N
here N=p

insert p for each ption and check for an int value only p=20 gives 18*180/20=int value
Hence D
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by sumanr84 » Mon Feb 08, 2010 9:45 am
Even 24 will give integer value,

(22 * 180) / 24 = 165

Q does not hold only a valid answer.
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by shashank.ism » Mon Feb 08, 2010 10:25 am
Each angle is 180(p-2)/p.

180-360/p=k So 360/p has to be an integer.
Factors of 360 = 23.32.51
So there are 4.3.2 solutions, but we exclude 1 and 2, because p>= 3 So , 24 -2 =22

Hence, C is the right answer
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by sumanr84 » Mon Feb 08, 2010 7:20 pm
shashank.ism wrote:Each angle is 180(p-2)/p.

180-360/p=k So 360/p has to be an integer.
Factors of 360 = 23.32.51
So there are 4.3.2 solutions, but we exclude 1 and 2, because p>= 3 So , 24 -2 =22

Hence, C is the right answer
Could not quite understand your logic after Factors of 360 ( something wrong with BTG editor here )

360 = 2 pow3 * 3 pow 2 * 5

How did you proceed after this point, pls explain ?

Also, how can you prove Ans D by thephoenix wrong ? It seems fine in substitution.
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