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If Polygon X has fewer than 9 sides, how many sides does

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If Polygon X has fewer than 9 sides, how many sides does [#permalink]

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11 Feb 2012, 18:15
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If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

(1) The sum of the interior angles of Polygon X is divisible by 16.

(2) The sum of the interior angles of Polygon X is divisible by 15.
[Reveal] Spoiler: OA

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11 Feb 2012, 18:28
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If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

Sum of inner angles of polygon=180*(x-2), where x is # of sides. Given x<9. Question x=?

(1) The sum of the interior angles of Polygon X is divisible by 16 --> 180*(x-2)=16k --> 45(x-2)=4k --> x-2 must be a multiple of 4 (as 45 is not) --> since x<9 then the only acceptable value of x is 6. Sufficient.

(2) The sum of the interior angles of Polygon X is divisible by 15 --> 180*(x-2)=15m --> 12(x-2)=m --> x can be any integer from 3 to 8, inclusive. Not sufficient. (We could even not consider this statement at all: as sum of inner angles of polygon is 180*(x-2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)

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Polygon with fewer than 9 sides [#permalink]

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03 Mar 2012, 15:03
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If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

(1) The sum of the interior angles of Polygon X is divisible by 16.

(2) The sum of the interior angles of Polygon X is divisible by 15.

Any idea how to solve this? I know the sum of interior angles of a polygon is 180(n-2). But still not getting the correct answer.
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Re: Polygon with fewer than 9 sides [#permalink]

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03 Mar 2012, 15:21
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enigma123 wrote:
If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

(1) The sum of the interior angles of Polygon X is divisible by 16.

(2) The sum of the interior angles of Polygon X is divisible by 15.

Any idea how to solve this? I know the sum of interior angles of a polygon is 180(n-2). But still not getting the correct answer.

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15 Oct 2012, 07:47
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Bunuel wrote:
If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

Sum of inner angles of polygon=180*(x-2), where x is # of sides. Given x<9. Question x=?

(1) The sum of the interior angles of Polygon X is divisible by 16 --> 180*(x-2)=16k --> 45(x-2)=4k --> x-2 must be a multiple of 4 (as 45 is not) --> since x<9 then the only acceptable value of x is 6. Sufficient.

(2) The sum of the interior angles of Polygon X is divisible by 15 --> 180*(x-2)=15m --> 12(x-2)=m --> x can be any integer from 3 to 8, inclusive. Not sufficient. (We could even not consider this statement at all: as sum of inner angles of polygon is 180*(x-2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)

Hi Bunuel - Could you please correct me! I got stuck in statement 2, after seeing your sol it makes more sense. But still want to know what's wrong in the below

Stat 1 -> 180*(x-2)=16k

x-2 = 16k/180
x-2 = 2^2 * k / 3^2 * 5 (after canceling out all the primes)
so for the min value of k = 3^2 * 5 we get x -2 = 2^2 and x= 6

As A can have other primes as well let assume k has another 2 in it, then (x-2) = 2^2 * 2 we get x=10 -> this is not possible as we are constrained by the question stem . So this is sufficient

Stat 2 -> 180*(x-2)=15m
x-2 = m/ 12
so the min value of m can be 12 hence the nos of sides become 14. this is not possible as per the question stem. how to proceed from there?

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15 Oct 2012, 07:56
Jp27 wrote:
Bunuel wrote:
If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

Sum of inner angles of polygon=180*(x-2), where x is # of sides. Given x<9. Question x=?

(1) The sum of the interior angles of Polygon X is divisible by 16 --> 180*(x-2)=16k --> 45(x-2)=4k --> x-2 must be a multiple of 4 (as 45 is not) --> since x<9 then the only acceptable value of x is 6. Sufficient.

(2) The sum of the interior angles of Polygon X is divisible by 15 --> 180*(x-2)=15m --> 12(x-2)=m --> x can be any integer from 3 to 8, inclusive. Not sufficient. (We could even not consider this statement at all: as sum of inner angles of polygon is 180*(x-2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)

Hi Bunuel - Could you please correct me! I got stuck in statement 2, after seeing your sol it makes more sense. But still want to know what's wrong in the below

Stat 1 -> 180*(x-2)=16k

x-2 = 16k/180
x-2 = 2^2 * k / 3^2 * 5 (after canceling out all the primes)
so for the min value of k = 3^2 * 5 we get x -2 = 2^2 and x= 6

As A can have other primes as well let assume k has another 2 in it, then (x-2) = 2^2 * 2 we get x=10 -> this is not possible as we are constrained by the question stem . So this is sufficient

Stat 2 -> 180*(x-2)=15m
x-2 = m/ 12
so the min value of m can be 12 hence the nos of sides become 14. this is not possible as per the question stem. how to proceed from there?

If m=12, then x-2=12/12=1 --> x=3.
If m=24, then x-2=24/12=2 --> x=4.
...

Hope it's clear.
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15 Oct 2012, 08:07
Bunuel wrote:
Jp27 wrote:
Bunuel wrote:
If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

Sum of inner angles of polygon=180*(x-2), where x is # of sides. Given x<9. Question x=?

(1) The sum of the interior angles of Polygon X is divisible by 16 --> 180*(x-2)=16k --> 45(x-2)=4k --> x-2 must be a multiple of 4 (as 45 is not) --> since x<9 then the only acceptable value of x is 6. Sufficient.

(2) The sum of the interior angles of Polygon X is divisible by 15 --> 180*(x-2)=15m --> 12(x-2)=m --> x can be any integer from 3 to 8, inclusive. Not sufficient. (We could even not consider this statement at all: as sum of inner angles of polygon is 180*(x-2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)

Hi Bunuel - Could you please correct me! I got stuck in statement 2, after seeing your sol it makes more sense. But still want to know what's wrong in the below

Stat 1 -> 180*(x-2)=16k

x-2 = 16k/180
x-2 = 2^2 * k / 3^2 * 5 (after canceling out all the primes)
so for the min value of k = 3^2 * 5 we get x -2 = 2^2 and x= 6

As A can have other primes as well let assume k has another 2 in it, then (x-2) = 2^2 * 2 we get x=10 -> this is not possible as we are constrained by the question stem . So this is sufficient

Stat 2 -> 180*(x-2)=15m
x-2 = m/ 12
so the min value of m can be 12 hence the nos of sides become 14. this is not possible as per the question stem. how to proceed from there?

If m=12, then x-2=12/12=1 --> x=3.
If m=24, then x-2=24/12=2 --> x=4.
...

Hope it's clear.

I so sorry for posting such a dumb question. I guess too much math today... all nos are appearing blurry now!

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Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]

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16 Nov 2012, 19:43
Is this regular polygon, Because I was informed that , we can use (2n-4)*90-Sum of interior angles for regular polygon.

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Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]

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15 Aug 2013, 03:33
Bumping for review and further discussion.
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Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]

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18 Sep 2013, 12:44
(1) SUFFICIENT: Using the relationship 180(n – 2) = (sum of interior angles), we could calculate the sum of the interior angles for all the polygons that have fewer than 9 sides. Just the first two are shown below; it would take too long to calculate all of the possibilities.

(2) INSUFFICIENT: Statement (2) tells us that the sum of the interior angles of Polygon X is divisible by 15. Therefore, the prime factorization of the sum of the interior angles will include 3 × 5. Following the same procedure as above, we realize that both 3 and 5 are included in the prime factorization of 180. As a result, every one of the possibilities can be divided by 15 regardless of the number of sides.

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Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]

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15 Nov 2013, 14:54
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calreg11 wrote:
If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

(1) The sum of the interior angles of Polygon X is divisible by 16.

(2) The sum of the interior angles of Polygon X is divisible by 15.

If you like the x-games or played tony hawk when you were a child you won't have trouble remembering that

3 sides - 180
4- 360
5-540
6-720
7-900

Statement 1, has to be divisible by 15 so by 8 and by 2 or 2^4. Only one that fits the bill is 720. Hence A is suff

Statement 2. Div by 15. More than 1 possible answer so Insuff

Hence A

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Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]

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16 Mar 2015, 04:55
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Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]

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26 Apr 2015, 10:52
Bunuel wrote:
If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

Sum of inner angles of polygon=180*(x-2), where x is # of sides. Given x<9. Question x=?

(1) The sum of the interior angles of Polygon X is divisible by 16 --> 180*(x-2)=16k --> 45(x-2)=4k --> x-2 must be a multiple of 4 (as 45 is not) --> since x<9 then the only acceptable value of x is 6. Sufficient.

(2) The sum of the interior angles of Polygon X is divisible by 15 --> 180*(x-2)=15m --> 12(x-2)=m --> x can be any integer from 3 to 8, inclusive. Not sufficient. (We could even not consider this statement at all: as sum of inner angles of polygon is 180*(x-2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)

I used a different method, I found that 16 is 2^4; 180 only has 2^2 as prime factors so (n-2) must be a factor of 4 (8 is too big since n would be 10). Is it correct to approach it this way? can I use this moving forward? Cheers

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If Polygon X has fewer than 9 sides, how many sides does [#permalink]

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26 Apr 2015, 11:17
tgubbay1 wrote:
Bunuel wrote:
If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

Sum of inner angles of polygon=180*(x-2), where x is # of sides. Given x<9. Question x=?

(1) The sum of the interior angles of Polygon X is divisible by 16 --> 180*(x-2)=16k --> 45(x-2)=4k --> x-2 must be a multiple of 4 (as 45 is not) --> since x<9 then the only acceptable value of x is 6. Sufficient.

(2) The sum of the interior angles of Polygon X is divisible by 15 --> 180*(x-2)=15m --> 12(x-2)=m --> x can be any integer from 3 to 8, inclusive. Not sufficient. (We could even not consider this statement at all: as sum of inner angles of polygon is 180*(x-2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)

I used a different method, I found that 16 is 2^4; 180 only has 2^2 as prime factors so (n-2) must be a factor of 4 (8 is too big since n would be 10). Is it correct to approach it this way? can I use this moving forward? Cheers

Hello tgubbay1

It's inherently the same approach which was given by Bunuel in second comment:
Bunuel wrote:
180*(x-2)=16k --> 45(x-2)=4k --> x-2 must be a multiple of 4

You definetely can use it for moving forward:
$$n-2 = 4k$$ so you should fine $$n$$ that will satisfactory for statement $$2 < n < 9$$
and this will be only variant: $$6$$
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Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]

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29 Mar 2016, 18:21
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calreg11 wrote:
If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

(1) The sum of the interior angles of Polygon X is divisible by 16.

(2) The sum of the interior angles of Polygon X is divisible by 15.

the sum of the interior angles of a polygon is:
180(n-2)
possible # of sides: 3,4,5,6,7,8

1. tells us that 180(n-2) is divisible by 16
or 45(n-2) is divisible by 4.
it is divisible only when n-2 is divisible by 4.
only way it can be true is if n=6, or polygon X has 6 sides.
any other option does not work. so 1 is sufficient.

2. the sum of interior angles is divisible by 15.
well..180 is divisible by 15, and so is 360. first one has 3 sides, second one has 4 sides.
since more than one option is possible, 2 alone is insufficient.

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Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]

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24 Jun 2017, 17:08
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Re: If Polygon X has fewer than 9 sides, how many sides does   [#permalink] 24 Jun 2017, 17:08
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