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M15-29

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Intern
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Re: M15-29  [#permalink]

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New post 27 Dec 2016, 19:18
I guess nobody has realized yet that S2 can be answered in the same way as S1, you just need to check if 1001 is divisible by 7.

111 x 1001 = 111,111
222 x 1001 = 222,222
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M15-29  [#permalink]

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New post 13 May 2017, 02:05
For statement 1) I tried two different integers by long division.
For statement 2) I tried 111,111 by long divison.

It might be more work, but is definitely manageable under 2 minutes ;)
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New post 17 Jun 2017, 01:29
I think this is a high-quality question. Please edit the question because there is a typo.
It should be "k-1 th digit" not "k-th digit"
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New post 17 Jun 2017, 04:35
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New post 15 Oct 2017, 03:00
Can you please explain how did you arrive at N=abc∗1000+abc
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New post 15 Oct 2017, 03:06
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New post 19 Dec 2017, 06:07
Hi,

Bunuel how did you get this?

N=abc∗1000+abc=abc∗1001

Pls explain
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New post 19 Dec 2017, 07:51
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New post 20 Dec 2017, 17:09
Bunuel wrote:
Official Solution:

Statements (1) and (2) by themselves are sufficient. S1 tells us that the last three digits of \(N\) are the same as the first three digits: \(N = abcabc\). Note that \(N = abc*1000 + abc = abc*1001\). As 1001 is divisible by 7, \(N\) is also divisible by 7.

Answer: D


Dear Bunuel

How do you prove statement 2 to be sufficient by itself?

As I understood, it simply says that all numbers are equal but where can I go from this info?

thanks
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New post 20 Dec 2017, 20:30
Mo2men wrote:
Bunuel wrote:
Official Solution:

Statements (1) and (2) by themselves are sufficient. S1 tells us that the last three digits of \(N\) are the same as the first three digits: \(N = abcabc\). Note that \(N = abc*1000 + abc = abc*1001\). As 1001 is divisible by 7, \(N\) is also divisible by 7.

Answer: D


Dear Bunuel

How do you prove statement 2 to be sufficient by itself?

As I understood, it simply says that all numbers are equal but where can I go from this info?

thanks


Every 6 digit integer with the same digits is divisible by 7: 111111, ..., 999999. It's given in the solution that ALL of them can be written as 1001*something ans 1001 itself IS divisible by 7.
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Re: M15-29 &nbs [#permalink] 20 Dec 2017, 20:30

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