Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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]

Show Tags

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.

Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]

Show Tags

16 Mar 2015, 04:55

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]

Show Tags

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*(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]

Show Tags

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*(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: \(6\) _________________

Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]

Show Tags

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(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.

gmatclubot

Re: If Polygon X has fewer than 9 sides, how many sides does
[#permalink]
29 Mar 2016, 18:21

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...