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Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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1
9 00:00

Difficulty:   95% (hard)

Question Stats: 40% (02:12) correct 60% (01:40) wrong based on 130 sessions

### HideShow timer Statistics 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) $$F(1) = F(2) = ... = F(6)$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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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.

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Intern  Joined: 14 Jan 2012
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and S2 sufficient because N=a*111111 where 111111 is divisible by 7?

I haven't memorized divisibility rule for 7 Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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Boycot wrote:
and S2 sufficient because N=a*111111 where 111111 is divisible by 7?

I haven't memorized divisibility rule for 7 You can find divisibility rules here: math-number-theory-88376.html
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Intern  Joined: 09 Dec 2013
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Quote:
(For example, F(4) is the value of the hundreds digit of N)?

Sorry if I am wrong, but I believe the example should be F(4) is the value of the thousands digit of N .
Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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martinnotceoyet wrote:
Quote:
(For example, F(4) is the value of the hundreds digit of N)?

Sorry if I am wrong, but I believe the example should be F(4) is the value of the thousands digit of N .

123,456
1 - HUNDRED THOUSANDS
2 - TEN THOUSANDS
3 - THOUSANDS
4 - HUNDREDS
5 - TENS
6 - UNITS
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Intern  Joined: 09 Dec 2013
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Omg, sure ! Thank you very much
Intern  Joined: 09 Aug 2014
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I think this question is good and not helpful.
Text seems confusing. F(k) is the value of the k−th digit. So F(4) means the value of the 4th digit which is thousands, right?
Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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IonutCZ wrote:
I think this question is good and not helpful.
Text seems confusing. F(k) is the value of the k−th digit. So F(4) means the value of the 4th digit which is thousands, right?

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Dear Bunuel

Could you please tell me what's the reasoning behind Statement 2? Is it the above-discussed fact with x*111'111? If that is indeed the case, how should one get that 111111 is divisible by 7 within the 2 mins given?:) Would be more than happy to get enlighted...

Thx vm, Dark Might
Intern  Joined: 15 Apr 2015
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I think this the explanation isn't clear enough, please elaborate.
Intern  Joined: 17 Oct 2015
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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. If someone could elaborate more... I just cant understand the reasons related at the explanations...
Current Student D
Joined: 12 Aug 2015
Posts: 2609
Schools: Boston U '20 (M)
GRE 1: Q169 V154 ### Show Tags

mestrec wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. If someone could elaborate more... I just cant understand the reasons related at the explanations...

I agree with you...
And More importantly i am not sure divisibility by 7 is really on the GMAT
maybe mikemcgarry can shed some light on whether we really need to do such questions or not ..

Regards
chirag
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Intern  B
Joined: 05 Nov 2012
Posts: 45

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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.

Hi Bunuel,

I am not sure if this part of the question is constructed properly. F(k) is the value of the k−th digit, (For example, F(4) is the value of the hundreds digit of N). Going by the statement you have given F (4) should be the value of the 4th digit right?. Also if I look at statement 1, F(1) = F(4) then here you have implied that the 1st digit = 4th digit. Therefore the question itself has clarity issues since how can F (K) stand for the kth digit as well as be the value for hundreds digit. Hope my question is clear.
Intern  B
Joined: 05 Nov 2012
Posts: 45

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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.

Hi Bunuel,

I am not sure if this part of the question is constructed properly. F(k) is the value of the k−th digit, (For example, F(4) is the value of the hundreds digit of N). Going by the statement you have given F (4) should be the value of the 4th digit right?. Also if I look at statement 1, F(1) = F(4) then here you have implied that the 1st digit = 4th digit. Therefore the question itself has clarity issues since how can F (K) stand for the kth digit as well as be the value for hundreds digit. Hope my question is clear.
Math Expert V
Joined: 02 Aug 2009
Posts: 7764

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nishatfarhat87 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.

Hi Bunuel,

I am not sure if this part of the question is constructed properly. F(k) is the value of the k−th digit, (For example, F(4) is the value of the hundreds digit of N). Going by the statement you have given F (4) should be the value of the 4th digit right?. Also if I look at statement 1, F(1) = F(4) then here you have implied that the 1st digit = 4th digit. Therefore the question itself has clarity issues since how can F (K) stand for the kth digit as well as be the value for hundreds digit. Hope my question is clear.

Hi ..
why cant be two digits have same value ..
example 123123..
here F(1) = F(4) = 1..
similarly other numbers can be seen as F(2) = F(5) = 2..
F(3) = F(6) = 3..
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Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34 GPA: 3.8
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I think this is a high-quality question and I agree with explanation.
Manager  B
Status: One Last Shot !!!
Joined: 04 May 2014
Posts: 229
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 630 Q44 V32 GMAT 2: 680 Q47 V35 ### Show Tags

1
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.

Bunuel, thanks for the explanation!
I think you missed the explanation for S2. How did we conclude that S2 is sufficient?

I just realised that every 6-digit number with all same digits is divisible by 7. But, are we supposed to know this or we can prove S2's sufficiency algebraically too?
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One Kudos for an everlasting piece of knowledge is not a bad deal at all... ------------------------------------------------------------------------------------------------------------------------
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Intern  Joined: 27 Oct 2015
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Hi Buenel,

Request you to kindly explain algebraically why Statement 2 is sufficient.

Many Thanks.
Retired Moderator V
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1344
Location: Viet Nam

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dsheth7 wrote:
Hi Buenel,

Request you to kindly explain algebraically why Statement 2 is sufficient.

Many Thanks.

State 2 is explained here
m15-184061.html#p1420417

DarkMight wrote:
Dear Bunuel

Could you please tell me what's the reasoning behind Statement 2? Is it the above-discussed fact with x*111'111? If that is indeed the case, how should one get that 111111 is divisible by 7 within the 2 mins given?:) Would be more than happy to get enlighted...

Thx vm, Dark Might

stonecold wrote:
mestrec wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. If someone could elaborate more... I just cant understand the reasons related at the explanations...

I agree with you...
And More importantly i am not sure divisibility by 7 is really on the GMAT
maybe mikemcgarry can shed some light on whether we really need to do such questions or not ..

Regards
chirag

In my opinion, in actual GMAT, we are not allowed to use calculator, so every calculation is done by ourselves.

In case that we face this question, I think we could still complete it under 2 mins.

In statement (1), we simply check whether 1001 is divisible by 7 or not.

In statement (2), we simply check whether 111,111 is divisible by 7 or not. And the division by 7 is not complex. I believe that anyone could make it in about 10 secs.

Here is how I make the calculation
Attachment:
Untitled.png

>> !!!

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_________________ Re: M15-29   [#permalink] 17 Dec 2016, 11:51

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# M15-29

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