Quote:
Originally Posted by STALWART
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Quantitative Question # 29
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Question:
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
Sorry , Replying late as usual .......
Answer -- 22
For any regular polygon having " p " number of sides , the sum of all the interior angles = ( p-2 ) * 180 = ( 180*p ) - 360 ......... (1)
Now the for finding out the value of each angle = [ (180*p) - 360 ] / p
==>> 180 - ( 360/p ) ......... (2)
For any Number N = (a^p) * ( b^q) * ( c^r ) ........
Where a , b , c , ...... are all prime numbars .
Number of all factors of N ( including 1 and the number itself ) = (p+1)(q+1)(r+1)......
Here 360 = (2^3) * (3^2) * (5^1)
So total number of factors = 4 * 3 * 2 = 24 ....... (3)
But out of them p=1,2 are not acceptable as then in those cases value of (2) will come in negative or 0 .
Or Just imagine , Can we have any polygon which have 1 or 2 number of side/sides ? Impossible !!
So required Number of solutions = 24 - 2 = 22
Answer = 22
..... Sayonara
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same slightly different
any angle will be given by (p-2)*180/p
now if 180 is multiple of p then we have integers
180 = 3^2*2^2*5 = 18 factors
but we can have case when we have exhausted all the 2's of 180 but as p-2 and p have a diff of 2 we can cancel there two
so take out 2^2 we are left with 3^2*5
which give 6 factors
so total 24
bhai log ab ye batau where i have counted more