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M03-20

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M03-20 [#permalink]

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New post 16 Sep 2014, 00:20
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A
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Re M03-20 [#permalink]

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New post 16 Sep 2014, 00:20
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 \(n\).


Answer: C
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Re: M03-20 [#permalink]

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New post 12 Nov 2015, 06:34
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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!

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Re: M03-20 [#permalink]

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New post 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.

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M03-20 [#permalink]

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New post 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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Re: M03-20 [#permalink]

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New post 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?

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Re: M03-20 [#permalink]

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New post 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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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