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# Consider a regular polygon of p sides. The number of values

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Consider a regular polygon of p sides. The number of values  [#permalink]

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21 May 2011, 02:33
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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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Consider a regular polygon of p sides. The number of values  [#permalink]

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17 Jan 2012, 05:31
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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*(p-2)/p. i. e. only 20 and 24 yield integer value for 180*(p-2)/p.
Can someone please check the answers.

Couple of things:
1. Sum of interior angles of a polygon is given by $$180(n-2)$$ where $$n$$ is the number of sides (for example the sum of interior angles of a triangle is $$180(3-2)=180$$ degrees and the sum of interior angles of a quadrilateral is $$180(4-2)=360$$ 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: $$\frac{180(n-2)}{n}$$ (for example each interior angle of an equilateral triangle is $$\frac{180(3-2)}{3}=60$$ degrees and each interior angles of a square is $$\frac{180(4-2)}{4}=90$$ degrees).
For more on polygons check: http://gmatclub.com/forum/math-polygons-87336.html

2. Finding the Number of Factors of an Integer

First 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: http://gmatclub.com/forum/math-number-theory-88376.html

BACK 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) $$\frac{180(p-2)}{p}$$ is an integer (or how many sided regular polygons exist which have interior angles equal to an integer).

Now, $$\frac{180(p-2)}{p}=180-\frac{360}{p}$$ 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 $$\frac{180(p-2)}{p}$$ is an integer.

Finally, as polygon cannot have 1 or 2 sides (p cannot be 1 or 2) then only 24-2=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  [#permalink]

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21 May 2011, 03:33
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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

(n-2)*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  [#permalink]

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21 May 2011, 04:09
3
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

(n-2)*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 24-2 = 22.
Not sure if it was a typo by Alchemist.
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Re: Polygon + Number properties  [#permalink]

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21 May 2011, 05:01
1
1
sudhir18n wrote:

I think the answer should be 22 and not 23.
same method as above . but p>=3
hence 24-2 = 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  [#permalink]

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30 May 2011, 05:43
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  [#permalink]

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30 May 2011, 10:09
1
1
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  [#permalink]

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22 Jun 2011, 19:16
1
Can somebody please explain the question and the answer. I am not clear with the question.
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Re: Polygon + Number properties  [#permalink]

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19 Dec 2011, 07:27
[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.....? 2-By 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  [#permalink]

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19 Dec 2011, 10:35
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*(p-2)/p. i. e. only 20 and 24 yield integer value for 180*(p-2)/p.
Can someone please check the answers.
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Joined: 10 Jul 2013
Posts: 279
Re: Polygon + Number properties  [#permalink]

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14 Aug 2013, 11:57
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*(p-2)/p. i. e. only 20 and 24 yield integer value for 180*(p-2)/p.
Can someone please check the answers.

Couple of things:
1. Sum of interior angles of a polygon is given by $$180(n-2)$$ where $$n$$ is the number of sides (for example the sum of interior angles of a triangle is $$180(3-2)=180$$ degrees and the sum of interior angles of a quadrilateral is $$180(4-2)=360$$ 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: $$\frac{180(n-2)}{n}$$ (for example each interior angle of an equilateral triangle is $$\frac{180(3-2)}{3}=60$$ degrees and each interior angles of a square is $$\frac{180(4-2)}{4}=90$$ degrees).
For more on polygons check: math-polygons-87336.html

2. Finding the Number of Factors of an Integer

First 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: math-number-theory-88376.html

BACK 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) $$\frac{180(p-2)}{p}$$ is an integer (or how many sided regular polygons exist which have interior angles equal to an integer).

Now, $$\frac{180(p-2)}{p}=180-\frac{360}{p}$$ 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 $$\frac{180(p-2)}{p}$$ is an integer.

Finally, as polygon can not have 1 or 2 sides (p can not be 1 or 2) then only 24-2=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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Joined: 22 Nov 2013
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Consider a regular polygon of p sides. The number of values  [#permalink]

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15 Dec 2014, 08:14
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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Posts: 17
Re: Consider a regular polygon of p sides. The number of values  [#permalink]

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05 Oct 2015, 11:28
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*(p-2)/p. i. e. only 20 and 24 yield integer value for 180*(p-2)/p.
Can someone please check the answers.

Couple of things:
1. Sum of interior angles of a polygon is given by $$180(n-2)$$ where $$n$$ is the number of sides (for example the sum of interior angles of a triangle is $$180(3-2)=180$$ degrees and the sum of interior angles of a quadrilateral is $$180(4-2)=360$$ 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: $$\frac{180(n-2)}{n}$$ (for example each interior angle of an equilateral triangle is $$\frac{180(3-2)}{3}=60$$ degrees and each interior angles of a square is $$\frac{180(4-2)}{4}=90$$ degrees).
For more on polygons check: math-polygons-87336.html

2. Finding the Number of Factors of an Integer

First 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: math-number-theory-88376.html

BACK 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) $$\frac{180(p-2)}{p}$$ is an integer (or how many sided regular polygons exist which have interior angles equal to an integer).

Now, $$\frac{180(p-2)}{p}=180-\frac{360}{p}$$ 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 $$\frac{180(p-2)}{p}$$ is an integer.

Finally, as polygon can not have 1 or 2 sides (p can not be 1 or 2) then only 24-2=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 24-1-3=20. Can you clarify me on this?
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Joined: 02 Sep 2009
Posts: 61403
Re: Consider a regular polygon of p sides. The number of values  [#permalink]

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06 Oct 2015, 03:48
1
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*(p-2)/p. i. e. only 20 and 24 yield integer value for 180*(p-2)/p.
Can someone please check the answers.

Couple of things:
1. Sum of interior angles of a polygon is given by $$180(n-2)$$ where $$n$$ is the number of sides (for example the sum of interior angles of a triangle is $$180(3-2)=180$$ degrees and the sum of interior angles of a quadrilateral is $$180(4-2)=360$$ 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: $$\frac{180(n-2)}{n}$$ (for example each interior angle of an equilateral triangle is $$\frac{180(3-2)}{3}=60$$ degrees and each interior angles of a square is $$\frac{180(4-2)}{4}=90$$ degrees).
For more on polygons check: math-polygons-87336.html

2. Finding the Number of Factors of an Integer

First 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: math-number-theory-88376.html

BACK 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) $$\frac{180(p-2)}{p}$$ is an integer (or how many sided regular polygons exist which have interior angles equal to an integer).

Now, $$\frac{180(p-2)}{p}=180-\frac{360}{p}$$ 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 $$\frac{180(p-2)}{p}$$ is an integer.

Finally, as polygon can not have 1 or 2 sides (p can not be 1 or 2) then only 24-2=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 24-1-3=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), $$\frac{180(p-2)}{p}$$ is an integer. But since p cannot be 1 or 2, then only 24-2=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  [#permalink]

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01 Aug 2016, 09:33
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]

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