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M03-20

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Bunuel
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M03-20 [#permalink] New post   16 Sep 2014, 00:20
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87% (00:33) correct 13% (01:08) wrong based on 708 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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C
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Bunuel
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Re M03-20 [#permalink] New post   16 Sep 2014, 00:20
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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


\(n=15! + 13 = 1*2*3*...12*13*14*15+13\). Since 13 is a factor of both \(15!\) and the additional 13 in the expression, we can conclude that 13 is a divisor of \(n\). In other words, we can factor out 13 from the expression to get \(n=13*(1*2*3*...12*14*15+1)\), hence 13 is definitely a divisor of \(n\).


Answer: C
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Re: M03-20 [#permalink] 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.
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piyuusshh
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Re: M03-20 [#permalink] 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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Bunuel
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Re: M03-20 [#permalink] New post   07 Aug 2017, 16:47
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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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banerjr1
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Re: M03-20 [#permalink] New post   25 Jun 2019, 07:29
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I solved it in the following method. It might be slightly more involved than Bunuel's solution.

\(n=15!+13\)
Let \(x\) be the divisor.

\(\frac{n}{x} = \frac{(15!+13)}{x}\)
\(\frac{n}{x} = \frac{15!}{x}+ \frac{13}{x}\)
The only value where \(\frac{n}{x}\) is an integer is where \(x = 13\).
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Re: M03-20 [#permalink] New post   12 Aug 2023, 10:49
I think this is a high-quality question and I agree with explanation.
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Bunuel
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Re: M03-20 [#permalink] New post   13 Aug 2023, 12:53
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M03-20 [#permalink]
13 Aug 2023, 12:53
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