It is currently 20 Jan 2018, 13:08

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M03-20

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139593 [0], given: 12794

### Show Tags

15 Sep 2014, 23:20
Expert's post
4
This post was
BOOKMARKED
00:00

Difficulty:

5% (low)

Question Stats:

87% (00:17) correct 13% (00:58) wrong based on 173 sessions

### HideShow timer Statistics

If $$n=15!+13$$, which of the following is a divisor of $$n$$?

A. 15
B. 14
C. 13
D. 7
E. 2
[Reveal] Spoiler: OA

_________________

Kudos [?]: 139593 [0], given: 12794

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139593 [0], given: 12794

### Show Tags

15 Sep 2014, 23:20
Expert's post
1
This post was
BOOKMARKED
Official Solution:

If $$n=15!+13$$, which of the following is a divisor of $$n$$?

A. 15
B. 14
C. 13
D. 7
E. 2

$$15!=1*2*3*...12*13*14*15$$, so $$n=1*2*3*...12*13*14*15+13$$. Factor out 13: $$n=13*(1*2*3*...12*14*15+1)$$, hence 13 is definitely a divisor of $$n$$.

_________________

Kudos [?]: 139593 [0], given: 12794

Manager
Joined: 02 May 2014
Posts: 116

Kudos [?]: 65 [1], given: 475

Schools: ESADE '16, HKU'16, SMU '16
GMAT 1: 620 Q46 V30

### Show Tags

12 Nov 2015, 05:34
1
KUDOS
Sum or difference of multiples of a number will also be multiple of the number.
Since, 15! and 13 are both multiples of 13, the result will also be a multiple of 13.
Thanks!

Kudos [?]: 65 [1], given: 475

Intern
Joined: 26 Sep 2016
Posts: 23

Kudos [?]: [0], given: 30

### Show Tags

15 Apr 2017, 10:15
Also from the concept of quotient and remainder: n is a multiple of certain numbers + remainder 13, hence for a number to be an integer when divided its remainder must be completely divisible and give 1 as a result.

Kudos [?]: [0], given: 30

Manager
Joined: 12 Jun 2016
Posts: 224

Kudos [?]: 49 [0], given: 152

Location: India
WE: Sales (Telecommunications)

### Show Tags

28 May 2017, 07:47
HI birsencal,

The Solution provided by Bunuel is by far the best and easiest to understand. I will just add few more points to what's already done make it clearer.

$$15!=1*2*3*...12*13*14*15$$, so
$$n=1*2*3*...12*13*14*15+13$$. Factor out 13:
$$n=13*(1*2*3*...12*14*15+1)$$, hence
$$n=13*(Positive \ integer \ N)$$, meaning
$$n= Multiple \ of \ 13$$

Means 13 can be a divisor of n

Scan the choices and to Pick C

Hope this helps!

birsencal wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. please elaborate. This solution is unclear to me. Thanks in advance!!!

_________________

My Best is yet to come!

Kudos [?]: 49 [0], given: 152

Intern
Joined: 13 Oct 2016
Posts: 2

Kudos [?]: [0], given: 1

### Show Tags

07 Aug 2017, 00:06
I used the concept of trailing zeroes as the last digits are ....0013 we can see that it is not divisible by any no. other than it self.
I am doubtfull whether it was mere luck in the question or is this way of doing it is also correct?

Kudos [?]: [0], given: 1

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139593 [0], given: 12794

### Show Tags

07 Aug 2017, 15:47
piyuusshh wrote:
I used the concept of trailing zeroes as the last digits are ....0013 we can see that it is not divisible by any no. other than it self.
I am doubtfull whether it was mere luck in the question or is this way of doing it is also correct?

15! + 13 = ...013 (actually it's 1,307,674,368,013)

Since the last digit is 3 (odd) we can get that it's not divisible by any even number (eliminate B and E) and not divisible by 5 because it does not end with 0 or 5 (eliminate A). But a number ending with 13 may or may not be divisible by 7 and may or may not be divisible by 13. For example, 413 is divisible by 7 and not divisible by 13.
_________________

Kudos [?]: 139593 [0], given: 12794

Re: M03-20   [#permalink] 07 Aug 2017, 15:47
Display posts from previous: Sort by

# M03-20

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.