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

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.

Target question: How many sides does Polygon X have?

Given: Polygon X has fewer than 9 sides

Useful rule: The sum of the angles in an n-sided polygon = (n - 2)(180º)

Since the polygon has FEWER than 9 sides, there are are exactly SIX possible cases:
case a: There are 8 sides, in which case the sum of the angles = (8 - 2)(180º) = 6(180º)
case b: There are 7 sides, in which case the sum of the angles = (7 - 2)(180º) = 5(180º)
case c: There are 6 sides, in which case the sum of the angles = (6 - 2)(180º) = 4(180º)
case d: There are 5 sides, in which case the sum of the angles = (5 - 2)(180º) = 3(180º)
case e: There are 4 sides, in which case the sum of the angles = (4 - 2)(180º) = 2(180º)
case f: There are 3 sides, in which case the sum of the angles = (3 - 2)(180º) = 180º

Statement 1: The sum of the interior angles of Polygon X is divisible by 16.
Only case c (6 sides) satisfies this condition.
4(180º) = 720, and 720 is divisible by 16.
Since no other cases satisfy the condition in statement 1, it MUST be the case that the polygon has 6 sides
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The sum of the interior angles of Polygon X is divisible by 15.
Since 180 is divisible by 15, we can be certain that any multiple of 180 is also divisible by 15.
So, cases a through to f all satisfy the condition in statement 2.
In other words, the polygon have have 8, 7, 6, 5, 4, or 3 sides
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

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!

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

Please somebody correct me.

(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.

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:


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: \(6\)

Hi Brent is there a faster way to solve the same ? I understood your approach but I guess it will take a few minutes before I could arrive at the answer . Also do you think such time consuming questions can feature on test day ?

I know that it LOOKS like my approach takes a lot of time but, on test day, I wouldn't write out all of those steps.
I might just create a small table with "# of sides" and "sum of angles"
We I have:
3: 180º
4: 360º
5: 540º
6: 720º
7: 900º
8: 1080º

Once we've created that table (20 seconds tops), it won't take long to deal with each statement.

Does that help?

Cheers,
Brent

\(\sum\nolimits_N { = \left( {N - 2} \right) \cdot 180\,\,\,\,\,\,\left[ {{\rm{degrees}}} \right]}\)

\(? = N\,\,\,\,\left( {3 \le \,\,N\,\,{\mathop{\rm int}} \,\, \le 8} \right)\)

\(\left( 1 \right)\,\,\,{\mathop{\rm int}} \,\, = \,\,{{\left( {N - 2} \right) \cdot 180} \over {16}}\,\, = \,\,{{\left( {N - 2} \right) \cdot 45} \over 4}\,\,\,\,\,\mathop \Rightarrow \limits^{GCF\left( {45,4} \right)\, = \,1} \,\,\,\,\,{{N - 2} \over 4} = {\mathop{\rm int}}\)

\(\left. \matrix{\\
3 \le \,\,N\,\,{\mathop{\rm int}} \,\, \le 8\,\, \hfill \cr \\
{{N - 2} \over 4} = {\mathop{\rm int}} \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,N = 6\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\)


\(\left( 2 \right)\,\,\,{\mathop{\rm int}} \,\, = \,\,{{\left( {N - 2} \right) \cdot 180} \over {15}}\,\, = \,\,\left( {N - 2} \right) \cdot 12\,\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,N = 3 \hfill \cr \\
\,{\rm{Take}}\,\,N = 4 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{INSUFF}}.\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Sum of interior angles of a polygon = (n-2)×180
Where n is the number of sides of a polygon
From first statement
180 is a multiple of 4, to make it a multiple of 16 only value of n that can be fitted is 6.
Sufficient
From second statement
180 is already a multiple of 15 so any value of n can be fitted
Insufficient
Option A is the answer

Posted from my mobile device

sum of interior angle of polygon ; 180*(n-2)

#1
The sum of the interior angles of Polygon X is divisible by 16.
180*(n-2)/16 ; 5*(n-2)/4 ;
n has to be 6 sufficient
#2
The sum of the interior angles of Polygon X is divisible by 15
180*(n-2)/15 ; 12*(n-2)
n can be any value from 1 to 8
insufficient
OPTION A is correct

180(n-2) = sum of interior angles of a polygon, where n = number of sides

We also know that n must be an integer and we can use this info to our advantage.

S1.
180(n-2) = 16 x Integer
90(n-2) = 8 x Integer
45(n-2) = 4 x Integer
45 is not divisible by 4, so n-2 must be div by 4. Acceptable value for n = 6 only, because 2<n<9
SUFFICIENT

S2.
180(n-2) = 15 x Integer
60(n-2) = 5 x Integer
60 is div by 5 so n-2 need not be div by 5
INSUFFICIENT