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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.
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Re: Polygon X [#permalink]
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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*(x2), 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*(x2)=16k > 45(x2)=4k > x2 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*(x2)=15m > 12(x2)=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*(x2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)Answer: A.
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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(n2). But still not getting the correct answer.
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Re: Polygon with fewer than 9 sides [#permalink]
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Re: Polygon X [#permalink]
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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*(x2), 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*(x2)=16k > 45(x2)=4k > x2 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*(x2)=15m > 12(x2)=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*(x2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)
Answer: A. 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*(x2)=16k x2 = 16k/180 x2 = 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 (x2) = 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*(x2)=15m x2 = 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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Re: Polygon X [#permalink]
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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*(x2), 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*(x2)=16k > 45(x2)=4k > x2 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*(x2)=15m > 12(x2)=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*(x2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)
Answer: A. 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*(x2)=16k x2 = 16k/180 x2 = 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 (x2) = 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*(x2)=15m x2 = 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 x2=12/12=1 > x=3. If m=24, then x2=24/12=2 > x=4. ... Hope it's clear.
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Re: Polygon X [#permalink]
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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*(x2), 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*(x2)=16k > 45(x2)=4k > x2 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*(x2)=15m > 12(x2)=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*(x2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)
Answer: A. 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*(x2)=16k x2 = 16k/180 x2 = 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 (x2) = 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*(x2)=15m x2 = 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 x2=12/12=1 > x=3. If m=24, then x2=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 (2n4)*90Sum of interior angles for regular polygon.
Please somebody correct me.



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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



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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.
The correct answer is A.



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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 xgames or played tony hawk when you were a child you won't have trouble remembering that 3 sides  180 4 360 5540 6720 7900 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 Cheers! J Kudos if you like!



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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*(x2), 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*(x2)=16k > 45(x2)=4k > x2 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*(x2)=15m > 12(x2)=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*(x2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)
Answer: A. I used a different method, I found that 16 is 2^4; 180 only has 2^2 as prime factors so (n2) 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*(x2), 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*(x2)=16k > 45(x2)=4k > x2 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*(x2)=15m > 12(x2)=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*(x2) and 180 is a multiple of 15, then all polygons will have the sum of the interior angles divisible by 15.)
Answer: A. I used a different method, I found that 16 is 2^4; 180 only has 2^2 as prime factors so (n2) 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 tgubbay1It's inherently the same approach which was given by Bunuel in second comment: Bunuel wrote: 180*(x2)=16k > 45(x2)=4k > x2 must be a multiple of 4 You definetely can use it for moving forward: \(n2 = 4k\) 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
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(n2) possible # of sides: 3,4,5,6,7,8 1. tells us that 180(n2) is divisible by 16 or 45(n2) is divisible by 4. it is divisible only when n2 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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