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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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20 Feb 2008, 07:22
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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: mgmat - prime factorization [#permalink]

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20 Feb 2008, 08:26
A.
sum of angles in polygon:
(n-2)*180

n<9

(n-2)*3^2* 2^2* 5

Statement II works for every n. insuff
Statement I will work if n = 6 as 6-2= 4. We need an extra 2^2 in our prime factorization to have the sum of angles divisible by 16.
if n = 6 --> (6-2)*3^2 *2^2 * 5
(4 *3^2 *4* 5)/16
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Re: mgmat - prime factorization [#permalink]

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20 Feb 2008, 08:29
bmwhype2 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 angles of any polygon = (#of sides - 2) * 180

Therefore,

6*180 = (2*3)(2*2*3*3*5)
5*180 = 5 (2*2*3*3*5)
4*180 = 2*2 (2*2*3*3*5)
3*180 = 3 (2*2*3*3*5)
2*180 = 2 (2*2*3*3*5)
1*180 = (2*2*3*3*5)

1) prime factors of 16 are 2*2*2*2 suff b/c only a hexagon has FOUR 2s in its prime factor set.

2) prime factors of 15 are 3*5 insuff, b/c all of the above polygons have these two prime factors.

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Re: mgmat - prime factorization [#permalink]

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20 Feb 2008, 10:13
A

hexagon.

Use (n-2)180

So check n = 3-8, ie 180 360 540 720 900 1080

180/16 = 11.25
360/16 = 22.5
540 = 33.75
720 = 45
900 = 56.25
1080 = 67.5

Hexagon.
Re: mgmat - prime factorization   [#permalink] 20 Feb 2008, 10:13
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# If Polygon X has fewer than 9 sides, how many sides does

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