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If n is a positive integer, and r is the remainder when

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If n is a positive integer, and r is the remainder when  [#permalink]

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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.

Originally posted by LM on 27 Apr 2010, 08:45.
Last edited by Bunuel on 14 Mar 2012, 04:48, edited 1 time in total.
Edited the question and added the OA
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If n is a positive integer, and r is the remainder when  [#permalink]

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New post 27 Apr 2010, 09:23
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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?

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.
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Re: GMAT PREP (DS)  [#permalink]

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New post 27 Apr 2010, 09:29
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]

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Re: GMAT PREP (DS)  [#permalink]

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New post 27 Jul 2010, 15:21
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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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Re: GMAT PREP (DS)  [#permalink]

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New post 27 Jul 2010, 15:40
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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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New post 30 Sep 2010, 21:37
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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: Remainder Problem  [#permalink]

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New post 30 Sep 2010, 21:45
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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New post 30 Sep 2010, 21:51
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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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New post 30 Sep 2010, 21:56
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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New post 01 Oct 2010, 09:25
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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New post 01 Oct 2010, 09:36
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 .... :-D
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Re: GMAT PREP (DS)  [#permalink]

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New post 01 Oct 2010, 09:50
I even replaced my facebook photo with this one ;)
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Re: GMAT PREP (DS)  [#permalink]

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New post 01 Oct 2010, 22:26
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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New post 01 Oct 2010, 23:05
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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New post 20 Oct 2010, 22:26
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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

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New post 14 Mar 2012, 04:31
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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If n is a positive integer, and r is the remainder when 4 + 7n i  [#permalink]

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New post 26 Oct 2012, 04:47
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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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New post 09 Aug 2013, 08:17
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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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Re: If n is a positive integer, and r is the remainder when  [#permalink]

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New post 12 Aug 2013, 10:31
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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: If n is a positive integer, and r is the remainder when &nbs [#permalink] 12 Aug 2013, 10:31

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