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M15#29

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powerka
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M15#29 [#permalink] New post   06 Oct 2009, 13:53
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:
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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.
The correct answer is D.


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powerka
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Re: M15#29 [#permalink] New post   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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diddygmat
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Re: M15#29 [#permalink] New post   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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kapilhede17
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Re: M15#29 [#permalink] New post   15 Mar 2013, 23:33
Can some one comment here , Since this figures in the gmatclub tests ?
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Re: M15#29 [#permalink] New post   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.

Please provide your comments on this.

Thanks.
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Re: M15#29 [#permalink] New post   04 Jun 2014, 03:31
Expert Reply
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.

Please provide your comments on this.

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
Moderator:
Bunuel
Math Expert
93007 posts

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