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# M15-29

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Intern
Joined: 26 Sep 2015
Posts: 4

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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
Intern
Joined: 07 Feb 2016
Posts: 21
GMAT 1: 650 Q47 V34
GMAT 2: 710 Q48 V39

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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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Joined: 04 Apr 2017
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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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Joined: 02 Sep 2009
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17 Jun 2017, 04:35
venkateshksep24 wrote:
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"

Why do you think that's it's a typo?
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Joined: 09 Apr 2017
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15 Oct 2017, 03:00
Can you please explain how did you arrive at N=abc∗1000+abc
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15 Oct 2017, 03:06
GeetikaR wrote:
Can you please explain how did you arrive at N=abc∗1000+abc

We can write 6-digit number abc,abc as abc∗1000+abc because abc,abc = abc,000 + abc = abc∗1000+abc. For example, 123,123 = 123,000 + 123 = 123*1000 + 123.
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Joined: 18 Mar 2015
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19 Dec 2017, 06:07
Hi,

Bunuel how did you get this?

N=abc∗1000+abc=abc∗1001

Pls explain
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Joined: 02 Sep 2009
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19 Dec 2017, 07:51
girishiyer wrote:
Hi,

Bunuel how did you get this?

N=abc∗1000+abc=abc∗1001

Pls explain

Factor our abc: abc∗1000 + abc = abc(1000 + 1) = abc∗1001.
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Joined: 26 Mar 2013
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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.

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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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.

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   [#permalink] 20 Dec 2017, 20:30

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# M15-29

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