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If n is a positive integer, and r is the remainder when [#permalink]
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27 Apr 2010, 09:45
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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.
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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]
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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 \(n=3k1\) > \(4+7(3k1)=3q+r\) > \(3(7k1q)=r\) > 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: Remainder Problem [#permalink]
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30 Sep 2010, 22:37
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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: GMAT PREP (DS) [#permalink]
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27 Jul 2010, 16:21
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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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If n is a positive integer, and r is the remainder when 4 + 7n i [#permalink]
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26 Oct 2012, 05:47
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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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Re: If n is a positive integer, and r is the remainder when [#permalink]
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12 Aug 2013, 11:31
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As n is +ve, so to fulfill the condition that (n+1) is divisible by 3, n can be 2, 5, 8, 11, 14 and for these values, when (4+7n) is divided by 3, it leaves remainder 0 everytime. so (1) is sufficient.
(2) n>20 is not required as for 0<n<20, we get the same remainder as for n>20.
Correct me if this method is wrong.



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Re: GMAT PREP (DS) [#permalink]
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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 \(n=3k1\) > \(4+7(3k1)=3q+r\) > \(3(7k1p)=r\) > 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]
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29 Apr 2010, 22:47



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Re: GMAT PREP (DS) [#permalink]
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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]
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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]
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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]
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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]
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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]
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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]
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01 Oct 2010, 10:50



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Re: GMAT PREP (DS) [#permalink]
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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]
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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]
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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) + (r4)/7 n + 1 = (3q/7) + (r4)/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., (r4/7) + 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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Re: If n is a positive integer, and r is the remainder when [#permalink]
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09 Aug 2013, 09:17
Q. What is r? (1). (n+1) div by 3 By question stem we know that 4 + 7n = 3Q + r Splitting the equation as below 4+4n+3n = 3Q + r => 3n + 4 (n+1) = 3Q + r LHS is divisible by 3 as (n+1) is div by 3, so RHS should also be divisible by 3 hence r should be 0 (2). 4 + 7n = 3Q +r Case 1: n=21 4 + 7*21 = 3Q +r LHS gives 4 as remainder when divided by 3 so r=4 Case 2: n=22 4 + 7*22 = 3Q + r No info about r, hence inconsistent (A) it is!
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