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If n is a positive integer, and r is the remainder when [#permalink]
 27 Apr 2010, 09:45
Question Stats:
62% (02:15) correct
37% (02:34) wrong based on 2 sessions
If n is a positive integer and r is the remainder when 4 + 7n is divided by 3, what is the value of r? (1) n+1 is divisible by 3 (2) n>20.
Last edited by Bunuel on 14 Mar 2012, 05:48, edited 1 time in total.
Edited the question and added the OA
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Re: GMAT PREP (DS) [#permalink]
27 Apr 2010, 10:23
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LM wrote: Please explain the answer...... If n is a positive integer, and r is the remainder when 4 + 7n is divided by 3, what is the value of r?r is the remainder when 4n+7 is divided by 3 --> 4+7n=3q+r, where r is an integer 0\leq{r}<3. r=?(1) n+1 is divisible by 3 --> n+1=3k, or n=3k-1 --> 4+7(3k-1)=3q+r --> 3(7k-1-q)=r --> so r is multiple of 3, but it's an integer in the range 0\leq{r}<3. Only multiple of 3 in this range is 0 --> r=0. Sufficient. (2) n>20. Clearly not sufficient. n=21, 4+7n=151=3q+r, r=1 BUT n=22, 4+7n=158=3q+r, r=2. Not sufficient. Answer: A. P.S. Please post DS questions in DS subforum.
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Re: GMAT PREP (DS) [#permalink]
27 Apr 2010, 10:29
Bunuel wrote: LM wrote: Please explain the answer...... If n is an integer, and r is the remainder when 4+7n is divided by 3, what is the value of r?r is the remainder when 4n+7 is divided by 3 --> 4+7n=3q+r, where r is an integer 0\leq{r}<3. r=?(1) n+1 is divisible by 3 --> n+1=3k, or n=3k-1 --> 4+7(3k-1)=3q+r --> 3(7k-1-p)=r --> so r is multiple of 3, but it's an integer in the range 0\leq{r}<3. Only multiple of 3 in this range is 0 --> r=0. Sufficient. (2) n>20. Clearly not sufficient. n=21, 4+7n=151=3q+r, r=1 BUT n=22, 4+7n=158=3q+r, r=2. Not sufficient. Answer: A. P.S. Please post DS questions in DS subforum. Thanks for detailed explanation. Sorry...in a haste I posted the DS questions in different forum. I will rectify this mistake next time onwards. Exam date is nearby ..so was in haste to get the explanations...thanks again.. You mentioned the 0\leq{r}<3 above, because remainder can't be more than the divisor. Is this correct?
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Re: GMAT PREP (DS) [#permalink]
29 Apr 2010, 22:47
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Re: GMAT PREP (DS) [#permalink]
27 Jul 2010, 16:21
i think i may have an easier way.... s1) 7n+4 = (6n+3)+(n+1) if (n+1)/3 = an integer, so must 3 times (n+1)....which is (6n+3) s2) Obviously NS
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Re: GMAT PREP (DS) [#permalink]
27 Jul 2010, 16:40
omarjmh wrote: i think i may have an easier way....
s1) 7n+4 = (6n+3)+(n+1)
if (n+1)/3 = an integer, so must 3 times (n+1)....which is (6n+3)
s2) Obviously NS Little correction: 3 times (n+1) is 3n+3 not (6n+3). But you are right, we can solve with this approach as well: (1) n+1 is divisible by 3 --> 7n+4=(4n+4)+3n=4(n+1)+3n --> 4(n+1) is divisible by 3 as n+1 is, and 3n is obviously divisible by 3 as it has 3 as multiple, thus their sum, 7n+4, is also divisible by 3, which means that remainder upon division 7n+4 by 3 will be 0. Sufficient. Hope it's clear.
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Re: Remainder Problem [#permalink]
30 Sep 2010, 22:37
Michmax3 wrote: If n is a positive integer and r is the remainder when 4+7n is divided by 3. What is the value of r?
1) n+1 is divisible by 3
2) n>20 7n+4=3(2n+1) + n+1 So remainder when divided by 3 will be same as remainder left by n+1 1) sufficient ... Gives the answer 2) insufficient ... Irrelevant. Eg n= 22,23,24 all are possible Answer is (a)
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Re: Remainder Problem [#permalink]
30 Sep 2010, 22:45
shrouded1 wrote: Michmax3 wrote: 7n+4=3(2n+1) + n+1
So remainder when divided by 3 will be same as remainder left by n+1
Can you explain how you get to 7n+4=3(2n+1)?
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Re: Remainder Problem [#permalink]
30 Sep 2010, 22:51
The information in the statement A is used favorably to tweak the equation in the question. Hence 7n+4 becomes 3(2n+1) + (n+1). Now since 3(2n+1) leaves a remainder 0 when divided by 3, the remainder of 7n+4 will be the same as the remainder of (n+1). Since (n+1) is also given in option A to be divisible by 3, hence remainder 0. Statement A is sufficient.
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Re: Remainder Problem [#permalink]
30 Sep 2010, 22:56
Michmax3 wrote: Can you explain how you get to 7n+4=3(2n+1)? Just trying to split it out into parts divisible by 3 7n becomes 6n+n 4 becomes 3+1
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Re: Remainder Problem [#permalink]
01 Oct 2010, 10:25
shrouded1 wrote: Michmax3 wrote: Can you explain how you get to 7n+4=3(2n+1)? Just trying to split it out into parts divisible by 3 7n becomes 6n+n 4 becomes 3+1 Thanks I see it now...btw is your avatar from the opening credits of Dexter?
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Re: Remainder Problem [#permalink]
01 Oct 2010, 10:36
Michmax3 wrote: shrouded1 wrote: Michmax3 wrote: Can you explain how you get to 7n+4=3(2n+1)? Just trying to split it out into parts divisible by 3 7n becomes 6n+n 4 becomes 3+1 Thanks I see it now...btw is your avatar from the opening credits of Dexter? YES !! It took me a lot of time to erase the credits which were in deep red painted right across the face and still maintain a semblance of originality in the image .... 
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Re: GMAT PREP (DS) [#permalink]
01 Oct 2010, 10:50
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Re: GMAT PREP (DS) [#permalink]
01 Oct 2010, 23:26
Profound identity question:
isnt "ezhilkumarank" the one with the dexter image?!! and not shrouded...
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Re: GMAT PREP (DS) [#permalink]
02 Oct 2010, 00:05
there is the cartoon dexter .... and then there is the serial killer dexter very different things !
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Re: GMAT PREP (DS) [#permalink]
20 Oct 2010, 23:26
Why am i getting negative Reminder here?
Given: 4 + 7n = 3q + r
If we simplify this further,
7n = 3q + r -4
n = (3q/7) + (r-4)/7
n + 1 = (3q/7) + (r-4)/7 + 1 ----- (1)
But From Statement 1, n+1 is divisible by 3.
So, (1) is divisible by 3 and hence the reminder is Zero
i.e., (r-4/7) + 1 = 0 ==> r = -3
Cheers!
Ravi
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1) n+1 = 3k and 7(n+1) = 7*3k so, 7n+7 is divisible by 3. this means that 7n+4 must also be divisible by 3.
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GMATPREP DATA SUFFICIENCY [#permalink]
08 Apr 2012, 02:43
please find the question attached. Please explain someone in detail. Attachment: 
question 2.JPG [ 53.08 KiB | Viewed 1134 times ]
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Re: GMATPREP DATA SUFFICIENCY [#permalink]
08 Apr 2012, 02:50
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If n is a positive integer, and r is the remainder when 4 + 7n i [#permalink]
26 Oct 2012, 05:47
A.
1) n+1 = 3 X (X is your Quotient) + 0(Remainder)
n+1=3X
Any Multiple of N+1 will be divisible by 3
Multiply by 7 --> 7(n+1)
So 7n+7 is divisible by 3 , implies 7n+4 is divisible by 3.
try nos for verification
2) n> 20 --- NS
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If n is a positive integer, and r is the remainder when 4 + 7n i
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26 Oct 2012, 05:47
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