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M15-29

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Math Expert
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15 Sep 2014, 23:56
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Difficulty:

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Question Stats:

40% (02:12) correct 60% (01:35) wrong based on 111 sessions

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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) $$F(1) = F(2) = ... = F(6)$$
[Reveal] Spoiler: OA

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15 Sep 2014, 23:56
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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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26 Sep 2014, 11:39
and S2 sufficient because N=a*111111 where 111111 is divisible by 7?

I haven't memorized divisibility rule for 7

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29 Sep 2014, 05:03
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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21 Oct 2014, 06:32
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 .

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21 Oct 2014, 06:35
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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21 Oct 2014, 07:02
Omg, sure ! Thank you very much

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19 Mar 2015, 04:18
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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19 Mar 2015, 04:24
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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30 Mar 2015, 07:39
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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

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27 Aug 2015, 09:48
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I think this the explanation isn't clear enough, please elaborate.

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10 Feb 2016, 03:54
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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...

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09 Mar 2016, 02:09
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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06 Apr 2016, 03:29
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.

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06 Apr 2016, 03:30
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.

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06 Apr 2016, 04:19
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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22 Aug 2016, 03:13
I think this is a high-quality question and I agree with explanation.

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24 Nov 2016, 07:10
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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17 Dec 2016, 03:33
Hi Buenel,

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

Many Thanks.

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17 Dec 2016, 10:51
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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