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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.)
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.)
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*(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 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.)
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*(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. ...
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.)
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*(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!
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
18 Sep 2013, 11:44
(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.
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
16 Mar 2015, 03:55
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Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
26 Apr 2015, 09:52
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.)
Answer: A.
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
If Polygon X has fewer than 9 sides, how many sides does [#permalink]
26 Apr 2015, 10:17
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.)
Answer: A.
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
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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