Last visit was: 21 May 2024, 17:08 It is currently 21 May 2024, 17:08
Toolkit
GMAT Club Daily Prep
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

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.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Manager
Joined: 27 Oct 2011
Posts: 85
Own Kudos [?]: 913 [46]
Given Kudos: 4
Location: United States
Concentration: Finance, Strategy
GPA: 3.7
WE:Account Management (Consumer Products)
Math Expert
Joined: 02 Sep 2009
Posts: 93373
Own Kudos [?]: 625635 [27]
Given Kudos: 81918
General Discussion
Manager
Joined: 22 Dec 2011
Posts: 175
Own Kudos [?]: 1048 [1]
Given Kudos: 32
Math Expert
Joined: 02 Sep 2009
Posts: 93373
Own Kudos [?]: 625635 [0]
Given Kudos: 81918
Re: Polygon X [#permalink]
Jp27 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.)

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.
Manager
Joined: 22 Dec 2011
Posts: 175
Own Kudos [?]: 1048 [0]
Given Kudos: 32
Re: Polygon X [#permalink]
Bunuel wrote:
Jp27 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.)

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!
Intern
Joined: 09 May 2012
Posts: 9
Own Kudos [?]: 18 [0]
Given Kudos: 7
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
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.
Intern
Joined: 09 Jul 2012
Posts: 6
Own Kudos [?]: 20 [0]
Given Kudos: 6
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
(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.
VP
Joined: 06 Sep 2013
Posts: 1343
Own Kudos [?]: 2401 [3]
Given Kudos: 355
Concentration: Finance
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
3
Kudos
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.

If you like the x-games or played tony hawk when you were a child you won't have trouble remembering that

3 sides - 180
4- 360
5-540
6-720
7-900

Statement 1, has to be divisible by 15 so by 8 and by 2 or 2^4. Only one that fits the bill is 720. Hence A is suff

Statement 2. Div by 15. More than 1 possible answer so Insuff

Hence A

Cheers!
J

Kudos if you like!
Intern
Joined: 13 Nov 2014
Posts: 28
Own Kudos [?]: 10 [0]
Given Kudos: 105
Location: France
Concentration: Healthcare, Statistics
WE:Consulting (Consulting)
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
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.)

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
Retired Moderator
Joined: 06 Jul 2014
Posts: 1010
Own Kudos [?]: 6355 [0]
Given Kudos: 178
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
If Polygon X has fewer than 9 sides, how many sides does [#permalink]
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.)

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:
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: $$6$$
Board of Directors
Joined: 17 Jul 2014
Posts: 2160
Own Kudos [?]: 1180 [3]
Given Kudos: 236
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
2
Kudos
1
Bookmarks
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.
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6817
Own Kudos [?]: 30289 [2]
Given Kudos: 799
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
2
Kudos
Top Contributor
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.

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

Cheers,
Brent
Manager
Joined: 03 Aug 2017
Posts: 76
Own Kudos [?]: 25 [0]
Given Kudos: 85
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
GMATPrepNow wrote:
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.

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

Cheers,
Brent

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 ?
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6817
Own Kudos [?]: 30289 [0]
Given Kudos: 799
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
Top Contributor
mimajit wrote:
GMATPrepNow wrote:
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.

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

Cheers,
Brent

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
Tutor
Joined: 12 Oct 2010
Status:GMATH founder
Posts: 892
Own Kudos [?]: 1382 [0]
Given Kudos: 56
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
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.

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

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

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

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

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

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

Regards,
Fabio.
Senior Manager
Joined: 18 Dec 2017
Posts: 270
Own Kudos [?]: 206 [0]
Given Kudos: 20
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
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
GMAT Club Legend
Joined: 18 Aug 2017
Status:You learn more from failure than from success.
Posts: 8027
Own Kudos [?]: 4126 [0]
Given Kudos: 242
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1:
545 Q79 V79 DI73
GPA: 4
WE:Marketing (Energy and Utilities)
Re: If Polygon X has fewer than 9 sides, how many sides does [#permalink]
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.

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
Senior Manager
Joined: 16 Oct 2020
Posts: 265
Own Kudos [?]: 163 [0]
Given Kudos: 2385
GMAT 1: 460 Q28 V26
GMAT 2: 550 Q39 V27
GMAT 3: 610 Q39 V35
GMAT 4: 650 Q42 V38
GMAT 5: 720 Q48 V41
If Polygon X has fewer than 9 sides, how many sides does [#permalink]
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
If Polygon X has fewer than 9 sides, how many sides does [#permalink]
Moderator:
Math Expert
93373 posts