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Consider a regular polygon of p sides. The number of values
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21 May 2011, 03:33
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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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Re: Polygon + Number properties
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17 Jan 2012, 06:31
maheshsrini wrote: Can somebody please explain the question and the answer. I am not clear with the question. BDSunDevil wrote: for a regular polygon of p sides .The number of values of p= 20 or 24, the polygon will have angles whose values in degrees can be expressed in integers. using the formula 180*(p2)/p. i. e. only 20 and 24 yield integer value for 180*(p2)/p. Can someone please check the answers. Couple of things: 1. Sum of interior angles of a polygon is given by \(180(n2)\) where \(n\) is the number of sides (for example the sum of interior angles of a triangle is \(180(32)=180\) degrees and the sum of interior angles of a quadrilateral is \(180(42)=360\) degrees). Question below talks about a regular polygon, which is a polygon with all equal sides and equal interior angles. Thus each interior angle of a regular polygon is given by: \(\frac{180(n2)}{n}\) (for example each interior angle of an equilateral triangle is \(\frac{180(32)}{3}=60\) degrees and each interior angles of a square is \(\frac{180(42)}{4}=90\) degrees). For more on polygons check: mathpolygons87336.html2. Finding the Number of Factors of an IntegerFirst make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers. The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\) Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors. For more on number properties check: mathnumbertheory88376.htmlBACK TO THE ORIGINAL 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 The question is: for how many values of \(p\) (where p is the # of sides of a regular polygon) \(\frac{180(p2)}{p}\) is an integer (or how many sided regular polygons exist which have interior angles equal to an integer). Now, \(\frac{180(p2)}{p}=180\frac{360}{p}\) to be an integer \(\frac{360}{p}\) must be an integer, so \(p\) must be a factor of 360. How many different positive factors does 360 have? Since \(360=2^3*3^2*5\) then # of factors is \((3+1)(2+1)(1+1)=24\), including 1 and 360. Thus if \(p\) is any of these 24 values (1, 2, 3, ... 360) then \(\frac{180(p2)}{p}\) is an integer. Finally, as polygon can not have 1 or 2 sides (p can not be 1 or 2) then only 242=22 regular polygons exist which have interior angles equal to an integer: 3 sided (equilateral triangle), 4 sided (square), 5 sided (regular pentagon), ..., 360 sided (trictohexacontagon ). Answer: C. Hope it's clear.
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Re: Polygon + Number properties
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21 May 2011, 04:33
hussi9 wrote: 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 formula (n2)*180/n so for angle to be an integer, 360/n should be an integer that reduces questions to " how many factors are there for 360" answer 24 ( three 2s, two 3s, one 5) correct answer is 23 as no polygon is possible with two sides




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Re: Polygon + Number properties
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21 May 2011, 05:09
Alchemist1320 wrote: hussi9 wrote: 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 formula (n2)*180/n so for angle to be an integer, 360/n should be an integer that reduces questions to " how many factors are there for 360" answer 24 ( three 2s, two 3s, one 5) correct answer is 23 as no polygon is possible with two sides I think the answer should be 22 and not 23. same method as above . but p>=3 hence 242 = 22. Not sure if it was a typo by Alchemist.



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Re: Polygon + Number properties
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21 May 2011, 06:01
sudhir18n wrote: I think the answer should be 22 and not 23. same method as above . but p>=3 hence 242 = 22. Not sure if it was a typo by Alchemist.
Opps I also over looked that... Its 22 and not 23 There cannot be polygon of 1 and 2 sides. 24  2 = 22
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Re: Polygon + Number properties
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30 May 2011, 06:43
I m sorry , but I didnt get the meaning of the below statement
"correct answer is 23 as no polygon is possible with two sides"
Could you please explain
thnks , Binu



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Re: Polygon + Number properties
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30 May 2011, 11:09
there are 24 possible polygons with given condition. This polygons count consider for polygons with sides 1 and 2. But polygons can have 3 or more sides . hence 24  2 = 22
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Re: Polygon + Number properties
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22 Jun 2011, 20:16
Can somebody please explain the question and the answer. I am not clear with the question.
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Re: Polygon + Number properties
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19 Dec 2011, 08:27
[quote="msbinu"]I m sorry , but I didnt get the meaning of the below statement
"correct answer is 23 as no polygon is possible with two sides"
Could you please explain
thnks ,
Hi 1. also do not understand the meaning of this.....? 2By the way, 360 is also divisible by 20, then why you picked 24 and not 20?
Can somebody explain?
thanks!!!



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Re: Polygon + Number properties
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19 Dec 2011, 11:35
for a regular polygon of p sides .The number of values of p= 20 or 24, the polygon will have angles whose values in degrees can be expressed in integers. using the formula 180*(p2)/p. i. e. only 20 and 24 yield integer value for 180*(p2)/p. Can someone please check the answers.



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Re: Polygon + Number properties
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14 Aug 2013, 12:57
Bunuel wrote: maheshsrini wrote: Can somebody please explain the question and the answer. I am not clear with the question. BDSunDevil wrote: for a regular polygon of p sides .The number of values of p= 20 or 24, the polygon will have angles whose values in degrees can be expressed in integers. using the formula 180*(p2)/p. i. e. only 20 and 24 yield integer value for 180*(p2)/p. Can someone please check the answers. Couple of things: 1. Sum of interior angles of a polygon is given by \(180(n2)\) where \(n\) is the number of sides (for example the sum of interior angles of a triangle is \(180(32)=180\) degrees and the sum of interior angles of a quadrilateral is \(180(42)=360\) degrees). Question below talks about a regular polygon, which is a polygon with all equal sides and equal interior angles. Thus each interior angle of a regular polygon is given by: \(\frac{180(n2)}{n}\) (for example each interior angle of an equilateral triangle is \(\frac{180(32)}{3}=60\) degrees and each interior angles of a square is \(\frac{180(42)}{4}=90\) degrees). For more on polygons check: mathpolygons87336.html2. Finding the Number of Factors of an IntegerFirst make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers. The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\) Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors. For more on number properties check: mathnumbertheory88376.htmlBACK TO THE ORIGINAL 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 The question is: for how many values of \(p\) (where p is the # of sides of a regular polygon) \(\frac{180(p2)}{p}\) is an integer (or how many sided regular polygons exist which have interior angles equal to an integer). Now, \(\frac{180(p2)}{p}=180\frac{360}{p}\) to be an integer \(\frac{360}{p}\) must be an integer, so \(p\) must be a factor of 360. How many different positive factors does 360 have? Since \(360=2^3*3^2*5\) then # of factors is \((3+1)(2+1)(1+1)=24\), including 1 and 360. Thus if \(p\) is any of these 24 values (1, 2, 3, ... 360) then \(\frac{180(p2)}{p}\) is an integer. Finally, as polygon can not have 1 or 2 sides (p can not be 1 or 2) then only 242=22 regular polygons exist which have interior angles equal to an integer: 3 sided (equilateral triangle), 4 sided (square), 5 sided (regular pentagon), ..., 360 sided. Answer: C. Hope it's clear. i wasn't sure about the 1 as one of the factors of 360 that's why answer was 23.... Now realize all the factorization includes 1 . really great thing i learned today..........
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Consider a regular polygon of p sides. The number of values
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15 Dec 2014, 09:14
I have a smaller solution... We don't need any geometry formula for this question.. . Now since we need integral interior angles ...so exterior angles must be integral too...!!(say interior angle = 60 degress than ext = 120..) So, If we find how many integral exterior angles are possible, we will also get all possible values of integral interior angles.
Now, Sum of all exterior angle = 360 degress..; so possible values of which divides 360 = (factors of 360)2 = 24 2 = 22 !!(2 because factors include 1 & 2 also and polygons cannot have n = 2) Kudos Please !!



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Re: Consider a regular polygon of p sides. The number of values
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05 Oct 2015, 12:28
Bunuel wrote: maheshsrini wrote: Can somebody please explain the question and the answer. I am not clear with the question. BDSunDevil wrote: for a regular polygon of p sides .The number of values of p= 20 or 24, the polygon will have angles whose values in degrees can be expressed in integers. using the formula 180*(p2)/p. i. e. only 20 and 24 yield integer value for 180*(p2)/p. Can someone please check the answers. Couple of things: 1. Sum of interior angles of a polygon is given by \(180(n2)\) where \(n\) is the number of sides (for example the sum of interior angles of a triangle is \(180(32)=180\) degrees and the sum of interior angles of a quadrilateral is \(180(42)=360\) degrees). Question below talks about a regular polygon, which is a polygon with all equal sides and equal interior angles. Thus each interior angle of a regular polygon is given by: \(\frac{180(n2)}{n}\) (for example each interior angle of an equilateral triangle is \(\frac{180(32)}{3}=60\) degrees and each interior angles of a square is \(\frac{180(42)}{4}=90\) degrees). For more on polygons check: mathpolygons87336.html2. Finding the Number of Factors of an IntegerFirst make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers. The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\) Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors. For more on number properties check: mathnumbertheory88376.htmlBACK TO THE ORIGINAL 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 The question is: for how many values of \(p\) (where p is the # of sides of a regular polygon) \(\frac{180(p2)}{p}\) is an integer (or how many sided regular polygons exist which have interior angles equal to an integer). Now, \(\frac{180(p2)}{p}=180\frac{360}{p}\) to be an integer \(\frac{360}{p}\) must be an integer, so \(p\) must be a factor of 360. How many different positive factors does 360 have? Since \(360=2^3*3^2*5\) then # of factors is \((3+1)(2+1)(1+1)=24\), including 1 and 360. Thus if \(p\) is any of these 24 values (1, 2, 3, ... 360) then \(\frac{180(p2)}{p}\) is an integer. Finally, as polygon can not have 1 or 2 sides (p can not be 1 or 2) then only 242=22 regular polygons exist which have interior angles equal to an integer: 3 sided (equilateral triangle), 4 sided (square), 5 sided (regular pentagon), ..., 360 sided (trictohexacontagon ). Answer: C. Hope it's clear. Hi Bunuel, Please enlighten me for that I think there are three 2s in the number of factors of 360. So I would think that 2413=20. Can you clarify me on this?



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Re: Consider a regular polygon of p sides. The number of values
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06 Oct 2015, 04:48
Bambaruush wrote: Bunuel wrote: maheshsrini wrote: Can somebody please explain the question and the answer. I am not clear with the question. BDSunDevil wrote: for a regular polygon of p sides .The number of values of p= 20 or 24, the polygon will have angles whose values in degrees can be expressed in integers. using the formula 180*(p2)/p. i. e. only 20 and 24 yield integer value for 180*(p2)/p. Can someone please check the answers. Couple of things: 1. Sum of interior angles of a polygon is given by \(180(n2)\) where \(n\) is the number of sides (for example the sum of interior angles of a triangle is \(180(32)=180\) degrees and the sum of interior angles of a quadrilateral is \(180(42)=360\) degrees). Question below talks about a regular polygon, which is a polygon with all equal sides and equal interior angles. Thus each interior angle of a regular polygon is given by: \(\frac{180(n2)}{n}\) (for example each interior angle of an equilateral triangle is \(\frac{180(32)}{3}=60\) degrees and each interior angles of a square is \(\frac{180(42)}{4}=90\) degrees). For more on polygons check: mathpolygons87336.html2. Finding the Number of Factors of an IntegerFirst make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers. The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\) Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors. For more on number properties check: mathnumbertheory88376.htmlBACK TO THE ORIGINAL 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 The question is: for how many values of \(p\) (where p is the # of sides of a regular polygon) \(\frac{180(p2)}{p}\) is an integer (or how many sided regular polygons exist which have interior angles equal to an integer). Now, \(\frac{180(p2)}{p}=180\frac{360}{p}\) to be an integer \(\frac{360}{p}\) must be an integer, so \(p\) must be a factor of 360. How many different positive factors does 360 have? Since \(360=2^3*3^2*5\) then # of factors is \((3+1)(2+1)(1+1)=24\), including 1 and 360. Thus if \(p\) is any of these 24 values (1, 2, 3, ... 360) then \(\frac{180(p2)}{p}\) is an integer. Finally, as polygon can not have 1 or 2 sides (p can not be 1 or 2) then only 242=22 regular polygons exist which have interior angles equal to an integer: 3 sided (equilateral triangle), 4 sided (square), 5 sided (regular pentagon), ..., 360 sided (trictohexacontagon ). Answer: C. Hope it's clear. Hi Bunuel, Please enlighten me for that I think there are three 2s in the number of factors of 360. So I would think that 2413=20. Can you clarify me on this? It does not matter how many 2'a re in 360. 360 has 24 factors. So, for 24 values of p (1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360), \(\frac{180(p2)}{p}\) is an integer. But since p cannot be 1 or 2, then only 242=22 regular polygons exist which have interior angles equal to an integer.
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Re: Consider a regular polygon of p sides. The number of values
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01 Aug 2016, 10:33
hussi9 wrote: 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 [/quote] Please find the solution
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Re: Consider a regular polygon of p sides. The number of values
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