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# M03-20

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Math Expert
Joined: 02 Sep 2009
Posts: 54440

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16 Sep 2014, 00:20
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Difficulty:

5% (low)

Question Stats:

87% (00:17) correct 13% (00:57) wrong based on 184 sessions

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

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

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Math Expert
Joined: 02 Sep 2009
Posts: 54440

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16 Sep 2014, 00:20
1
1
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 $$n$$.

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Joined: 02 May 2014
Posts: 94
Schools: ESADE '16, HKU'16, SMU '16
GMAT 1: 620 Q46 V30

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12 Nov 2015, 06:34
2
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!
Intern
Joined: 26 Sep 2016
Posts: 23

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15 Apr 2017, 11:15
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.
Manager
Joined: 12 Jun 2016
Posts: 214
Location: India
WE: Sales (Telecommunications)

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28 May 2017, 08:47
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!!!

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Intern
Joined: 14 Oct 2016
Posts: 1

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07 Aug 2017, 01:06
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?
Math Expert
Joined: 02 Sep 2009
Posts: 54440

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07 Aug 2017, 16:47
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.
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Re: M03-20   [#permalink] 07 Aug 2017, 16:47
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# M03-20

Moderators: chetan2u, Bunuel

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