A says that 180(n-2) is evenly divisible by 16, which would mean n has to be 6 (less than 9).
For B, n could be any value less than 9 and satisfy as 15 would divide 180.
Polygon X
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earth@work
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same as kris610, answer Apbanavara wrote:I think it's C .. not sure - Combining the 2 the only option is n=6
Because from A n can be 4 as well.
- pradeep
n cannot be 4, 180(n-2)/16=45(n-2)/4 now n-2 shud be divisible by 4 so smallest value of n=6 <9
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pbanavara
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Agree - my bad .. calculation mistakeearth@work wrote:same as kris610, answer Apbanavara wrote:I think it's C .. not sure - Combining the 2 the only option is n=6
Because from A n can be 4 as well.
- pradeep
n cannot be 4, 180(n-2)/16=45(n-2)/4 now n-2 shud be divisible by 4 so smallest value of n=6 <9
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4meonly
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My attempt:
sum of angles = 180*(n-2)
I factorized 180 = 2^2 * 3^2 * 5 *(n-2)
n<9
(1) The sum of the interior angles of Polygon X is divisible by 16.
16 = 2^4
The only way to make 2^2*3^2*5*(n-2) divisible by 16 (n-2) should give additional 2^2. because n<9 the only n is 6, because (6-2)=2^2
So the only sum is 2^2*3^2*5*(6-2) = 2^4*3^2*5 that is div by 16
SUFF
(2) The sum of the interior angles of Polygon X is divisible by 15.
With the same logic 180 is already divisible by 15 so ANY n will give thesum that is divisible by 15
INSUFF
A
sum of angles = 180*(n-2)
I factorized 180 = 2^2 * 3^2 * 5 *(n-2)
n<9
(1) The sum of the interior angles of Polygon X is divisible by 16.
16 = 2^4
The only way to make 2^2*3^2*5*(n-2) divisible by 16 (n-2) should give additional 2^2. because n<9 the only n is 6, because (6-2)=2^2
So the only sum is 2^2*3^2*5*(6-2) = 2^4*3^2*5 that is div by 16
SUFF
(2) The sum of the interior angles of Polygon X is divisible by 15.
With the same logic 180 is already divisible by 15 so ANY n will give thesum that is divisible by 15
INSUFF
A
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Stuart@KaplanGMAT
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First, let's jot down the key formula:logitech wrote:If 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.
# of interior degrees of an n sided polygon = 180(n-2)
In DS, we can start with whichever statement is easier. In this case, that's certainly (2).
(2) Well, 180 is divisible by 15. Therefore, all polygons will have interior angles divisible by 15.
Not only is this insufficient, it's completely useless information. Therefore, we can eliminate (b), (c) and (d). (Remember, (c) is only correct if each statement adds something that you need.)
(1) 180(n-2)/16 is an integer.
Factoring out:
45(n-2)/4 is an integer. Well, we can't get "4" out of "45", so 4 MUST be a factor of (n-2). The only integer between 3 and 8 that gives us what we need is 6: sufficient.
(1) is sufficient, (2) isn't: choose (A).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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