M03Q20 : Retired Discussions [Locked]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 04:16

### 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

# M03Q20

Author Message
Intern
Joined: 16 Apr 2007
Posts: 1
Followers: 0

Kudos [?]: 2 [1] , given: 0

### Show Tags

28 Apr 2009, 14:53
1
KUDOS
2
This post was
BOOKMARKED
Which of the following integers is a divisor of $$(15!+13)$$?

(A) 15
(B) 14
(C) 13
(D) 7
(E) 2

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Factorials can be rewritten as follows:
n! = n*(n-1)*(n-2)*(n-3)*...*(n-k)!

Where k can be any number from 1 to n. We can rewrite (15!+13) as follows:
15!+13 = (15*14*13*12!)+13 = 13*(15*14*12!+1)

Therefore, (15!+13) in its final form has a divisor of 13 in it but not 15, 14, 7, or 2.

Senior Manager
Joined: 01 Mar 2009
Posts: 372
Location: PDX
Followers: 6

Kudos [?]: 89 [0], given: 24

### Show Tags

29 Apr 2009, 20:18
13 ( 15*14*12*....1 + 1) can be compared to something like

x(y+1) is divisible by x and not y .So x is a factor of x(y+1) and not y ..
_________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

Manager
Joined: 22 Aug 2009
Posts: 99
Location: UK
Followers: 2

Kudos [?]: 16 [0], given: 25

### Show Tags

01 Feb 2010, 19:14
Alternate Explanation :
15 ! will be divided by even ( because it contains atleast one even)

15! + 13 becomes odd, so strike out (B) and (E)

Also 15 ! will be divided by 15 , so 15 ! + 13 will be 2 short of 15
similarly 7 can be eliminated.
_________________

FEB 15 2010 !!

well I would not disturb you after the D-day ..so please !!!

It Will Snow In Green , One Day !!!!

Intern
Joined: 25 Feb 2009
Posts: 13
Location: Nairobi
Schools: HBS
Followers: 3

Kudos [?]: 15 [5] , given: 0

### Show Tags

18 May 2010, 06:51
5
KUDOS
Which of the following integers is a divisor of (15!+13)?

(A) 15
(B) 14
(C) 13
(D) 7
(E) 2

C

explanation

(15!+13) = ((12!*13*14*15)+13) = 13 ((12!*14*15)+1)
Current Student
Joined: 13 Nov 2008
Posts: 8
Location: United States (CA)
Followers: 0

Kudos [?]: 15 [6] , given: 6

### Show Tags

19 May 2010, 05:48
6
KUDOS
The solution is pretty straightforward...!!!

The question asks if (15! + 13) divisible by any of the four choices. I can rewrite this as (a+b)/c = a/c + b/c. Here a = 15!, b=13 and c = one of the four answer choices.

We know 15! will contains all integers from 15 through 1----> so 15! is divisible by all choices.

Now 13 is divisible by only 13 and 1 because it is a prime number.

Therefore, 13 divides both 15! and obviously 13------> HENCE the answer.

Took me about 10 seconds to figure this out.
Manager
Joined: 28 Oct 2009
Posts: 92
Followers: 1

Kudos [?]: 104 [0], given: 42

### Show Tags

19 May 2010, 09:56
The solution is pretty straightforward...!!!

The question asks if (15! + 13) divisible by any of the four choices. I can rewrite this as (a+b)/c = a/c + b/c. Here a = 15!, b=13 and c = one of the four answer choices.

We know 15! will contains all integers from 15 through 1----> so 15! is divisible by all choices.

Now 13 is divisible by only 13 and 1 because it is a prime number.

Therefore, 13 divides both 15! and obviously 13------> HENCE the answer.

Took me about 10 seconds to figure this out.

nice
Current Student
Joined: 08 Jan 2009
Posts: 326
GMAT 1: 770 Q50 V46
Followers: 25

Kudos [?]: 136 [0], given: 7

### Show Tags

20 May 2011, 20:28
DLiu1214 wrote:
Hi,

Sorry to sound elementary but I am not familiar with this simplification:

15! + 13 = 13(15*14*12*... + 1)

Can someone explain the general form?

Thank you.

Not really sure what you mean by "general form". The simplification is just the distributive property over additive numbers

$$(x*c + y*c) = (x + y) * c$$

So:

$$15! + 13 = (15 * 14 * 13 * ... * 2 * 1) + 13 = (15 * 14 * 12 * 11 * ... * 2 * 1) * 13 + 1 * 13 = ((15 * 14 * 12 * 11 * ... * 2 * 1) + 1) * 13$$
Intern
Status: GMAT once done, Going for GMAT 2nd time.
Joined: 29 Mar 2010
Posts: 34
Location: Los Angeles, CA
Schools: Anderson, Haas, Ross, Kellog, Booth, McCombs
Followers: 0

Kudos [?]: 2 [0], given: 3

### Show Tags

22 May 2011, 16:48
i go with 13.

Reason: 15!/x + 13/x, what is x?
Since 13 is prime it can be divided by 13 and 1. Since 1 not an option and 15! has 13, x=13.
_________________

BP

Manager
Joined: 16 Sep 2010
Posts: 223
Location: United States
Concentration: Finance, Real Estate
GMAT 1: 740 Q48 V42
Followers: 9

Kudos [?]: 112 [0], given: 2

### Show Tags

30 May 2011, 20:24
While you can use math to explain it simple logic will also suffice.

Talking 15! automatically makes the end result divisible by all the components (so 15! is divisible by 13). So merely adding 13 to this end results guarantees to make it a multiple of 13.

I found it to be really easy (sub 15 sec question) when looked at in that light.
Manager
Status: And the Prep starts again...
Joined: 03 Aug 2010
Posts: 138
Followers: 2

Kudos [?]: 47 [0], given: 20

### Show Tags

03 Mar 2012, 19:09
analyst218 wrote:
15! has three zero's in it. therefore adding 13 to the number would be divisible only by 13.

Isn't this supposed to be the easiest solution? Experts let me know if this method is wrong.

Will have 3 zeroes because we have 3 combinations that can give us a zero 5*2, 10, 15*12
_________________

My First Blog on my GMAT Journey

Arise, Awake and Stop not till the goal is reached

Intern
Joined: 21 Feb 2012
Posts: 12
Location: Chile
Followers: 0

Kudos [?]: 6 [0], given: 3

### Show Tags

23 Mar 2012, 11:43
Why the zeros could be relevant in this case?

For instance,
$$12*13 - 13 = 13*(12 - 1)$$ is divisible by 13 (no zeros!).
$$12*13*10 - 13 = 13*(12*10 - 1)$$ is divisible by 13.
$$12*13*100 - 13 = 13*(12*100 - 1)$$ is divisible by 13.

$$(15! - 13)$$ is divisible by 13 because, although obvious, it is a multiple of 13. You can know that after performing the factorization $$13*(\frac{15!}{13} - 1)$$

Is $$254000 + 13$$ divisible by 13?
_________________

Francisco

Intern
Status: Preparing again for second attempt....
Joined: 11 Dec 2010
Posts: 24
WE 1: 6 years
Followers: 0

Kudos [?]: 9 [0], given: 2

### Show Tags

25 May 2012, 02:45
This question is brilliant as it tests the two concepts of remainder theorm -

1. (Multiplication Rule): This rule is partially applicable here, i would still write it for people who dont know it yet. In an expression (a*b*c)/d, if any of the individual members of the numerator is divisible by d, then the whole numerator is divisible by d. In other words, the remainder of the expression in such a case is zero.

2. (Addition Rule): This one is applicable here. Again in an expression (a+b+c)/d, the remainder will be zero if a, b and c, each, when individually divided by d, gives a remainder zero OR when these individual remainder add up to d itself. Lets take an example to understand better. We all know that 27 is divisible by 9. If i were to write 27 as 9+9+9, then each of the 9's leaves a remainder zero. However if i were to write 27 as 21+2+4, then 21, 2 and 4, when divided by 9, will leave a remainder of 3, 2 and 4 respectively. These remainders add up to 9.

Now lets look at the question: 13! + 13 is always going to be divisible by 13. Why? Apply the fundamentals given above. 13! is nothing but 13*12*11*10 ..... *3*2*1. This is divisible by 13 (Use Multiplication rule on 13! part).

13, the second half, is only divisible by 13 & 1 (It's a prime number). All answer choices EXCEPT 13, leave some remainder. ONLY 13 divides both parts of the expression, leaving an overall remainder zero.

I hope my explanation helped.

Intern
Joined: 24 Apr 2012
Posts: 3
GMAT Date: 08-08-2012
Followers: 0

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

### Show Tags

25 May 2012, 06:45
13 is a multiple of 13.
15! is a multiple of the numbers from 1 to 15 .Hence a multiple of 13.

The sum of multiples of a number is also a multiple of the number.
Since 13 is there in the option, no need to check for any other options.
Senior Manager
Joined: 21 Jan 2010
Posts: 344
Followers: 2

Kudos [?]: 176 [0], given: 12

### Show Tags

30 May 2012, 16:31
15! will be divided by every integer till 15 , the question is what integer divides 13, which is 13 itself. Hence C.
Intern
Joined: 21 May 2013
Posts: 7
Location: United States
GMAT 1: 640 Q49 V27
Followers: 0

Kudos [?]: 5 [1] , given: 15

### Show Tags

24 May 2013, 02:48
1
KUDOS
hey
the 15! = (15*14*13*12*11*10!)
so it is clear that 13 is a divisor of 15!, and adding 13 to it makes the quotient increase by 1
thus 13 is a divisor of (15!+13) .

if you like it, please give kuddos
Math Expert
Joined: 02 Sep 2009
Posts: 36531
Followers: 7070

Kudos [?]: 92960 [0], given: 10541

### Show Tags

24 May 2013, 02:57
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 $$n$$.

_________________
Intern
Joined: 18 May 2013
Posts: 6
GMAT 1: 770 Q51 V42
Followers: 0

Kudos [?]: 6 [0], given: 3

### Show Tags

24 May 2013, 04:12
!15 and 13 , both have 13 as their factors, so 13 is the answer
_________________

Find Your Own Dream, Make it Happen !

Intern
Joined: 06 May 2009
Posts: 11
Followers: 0

Kudos [?]: 1 [0], given: 3

### Show Tags

24 May 2013, 05:27
shakeandbake wrote:
Which of the following integers is a divisor of $$(15!+13)$$?

(A) 15
(B) 14
(C) 13
(D) 7
(E) 2

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Factorials can be rewritten as follows:
n! = n*(n-1)*(n-2)*(n-3)*...*(n-k)!

Where k can be any number from 1 to n. We can rewrite (15!+13) as follows:
15!+13 = (15*14*13*12!)+13 = 13*(15*14*12!+1)

Therefore, (15!+13) in its final form has a divisor of 13 in it but not 15, 14, 7, or 2.

(15!= 15x14x13x12x11x.......x2x1) is a multiple of 13 . When you add 13 to a multiple of 13, it is still a multiple of 13 and divisible by 13. So answer is C.
Intern
Joined: 08 Mar 2013
Posts: 19
Followers: 0

Kudos [?]: 3 [0], given: 7

### Show Tags

27 May 2013, 20:28
Just curious, would something like this be a 600-level question?
Math Expert
Joined: 02 Sep 2009
Posts: 36531
Followers: 7070

Kudos [?]: 92960 [1] , given: 10541

### Show Tags

27 May 2013, 23:45
1
KUDOS
Expert's post
Dixon wrote:
Just curious, would something like this be a 600-level question?

Yes, I'd say it's a sub-600 question.
_________________
Re: M03Q20   [#permalink] 27 May 2013, 23:45

Go to page    1   2    Next  [ 22 posts ]

Display posts from previous: Sort by

# M03Q20

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO 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®.