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# M15#29

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Manager
Joined: 22 Jul 2009
Posts: 191

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06 Oct 2009, 13:53
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If in a six-digit integer $$N$$, $$F(k)$$ is the value of the $$k-th$$ digit, is $$N$$ divisible by 7 (For example, $$F(4)$$ is the value of the hundreds digit of $$N$$ )?

(1) $$F(1) = F(4), F(2) = F(5), F(3) = F(6)$$
(2) $$F(1) = F(2) = ... = F(6)$$

(C) 2008 GMAT Club - m15#29

SOLUTION:
[Reveal] Spoiler:
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.

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Manager
Joined: 22 Jul 2009
Posts: 191

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06 Oct 2009, 13:59
I believe there is an error in this question.

If, as it says, F(4) is the value of the hundreds digit, then F(1) is the value of the decimal digit of N.

Given that N is an integer, then F(1)=0.

F(1)=F(2)=F(3)=F(4)=F(5)=F(6)=0 --> N=0 and it is not a 6 digit integer. This contradicts the question stem.

Or am I missing something?
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Intern
Joined: 22 Sep 2010
Posts: 9

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11 Jan 2012, 04:32
This question somehow doesnt make sense. How can F(4) be hundreds digit? shouldnt it 1000's digit..

Help will be really appreciated!
Thanks!

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Current Student
Joined: 14 May 2012
Posts: 75

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Location: United States
Concentration: Finance, Strategy
GMAT 1: 710 Q49 V38
GPA: 3.8
WE: Corporate Finance (Venture Capital)

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15 Mar 2013, 23:33
Can some one comment here , Since this figures in the gmatclub tests ?

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Manager
Joined: 10 Mar 2014
Posts: 238

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03 Jun 2014, 21:27
powerka wrote:
I believe there is an error in this question.

If, as it says, F(4) is the value of the hundreds digit, then F(1) is the value of the decimal digit of N.

Given that N is an integer, then F(1)=0.

F(1)=F(2)=F(3)=F(4)=F(5)=F(6)=0 --> N=0 and it is not a 6 digit integer. This contradicts the question stem.

Or am I missing something?

Hi Bunnel,

I also have a doubt here. if f(4) is hundreads digit as given in question then how f(1) can be units digit.

Thanks.

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Math Expert
Joined: 02 Sep 2009
Posts: 41872

Kudos [?]: 128639 [0], given: 12181

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04 Jun 2014, 03:31
PathFinder007 wrote:
powerka wrote:
If in a six-digit integer $$N$$, $$F(k)$$ is the value of the $$k-th$$ digit, is $$N$$ divisible by 7 (For example, $$F(4)$$ is the value of the hundreds digit of $$N$$ )?

(1) $$F(1) = F(4), F(2) = F(5), F(3) = F(6)$$
(2) $$F(1) = F(2) = ... = F(6)$$

I believe there is an error in this question.

If, as it says, F(4) is the value of the hundreds digit, then F(1) is the value of the decimal digit of N.

Given that N is an integer, then F(1)=0.

F(1)=F(2)=F(3)=F(4)=F(5)=F(6)=0 --> N=0 and it is not a 6 digit integer. This contradicts the question stem.

Or am I missing something?

Hi Bunnel,

I also have a doubt here. if f(4) is hundreads digit as given in question then how f(1) can be units digit.

Thanks.

F(1) is the value of HUNDRED THOUSANDS digit not the UNITS digit.

123,456
1 - HUNDRED THOUSANDS
2 - TEN THOUSANDS
3 - THOUSANDS
4 - HUNDREDS
5 - TENS
6 - UNITS

Hope it's clear.
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Re: M15#29   [#permalink] 04 Jun 2014, 03:31
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# M15#29

Moderator: Bunuel

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