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If n is a positive integer and r is the remainder when 4 + 7n is divid
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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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Originally posted by LM on 27 Apr 2010, 09:45.
Last edited by Bunuel on 17 Jun 2019, 04:07, edited 2 times in total.
Edited the question and added the OA




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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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27 Apr 2010, 10:23
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.
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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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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: If n is a positive integer and r is the remainder when 4 + 7n is divid
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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: If n is a positive integer and r is the remainder when 4 + 7n is divid
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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: If n is a positive integer and r is the remainder when 4 + 7n is divid
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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: If n is a positive integer and r is the remainder when 4 + 7n is divid
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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: If n is a positive integer and r is the remainder when 4 + 7n is divid
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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: If n is a positive integer and r is the remainder when 4 + 7n is divid
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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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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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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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Re: If n is a positive integer and r is the remainder when 4+7n
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27 Jul 2013, 12:52
forgmat 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
Please explain statmnt 1: n+1 ==>div by 3 therefore 7(n+1) ==>div. by 3 (7n+4)+3 ==>div by ===>this means 7n+4 is div by 3 sufficient statement 2: n>20 let n=30==>7n+4=214==>when divided by 3 remainder i s1 let n=40==>7n+4=284==>when divided by 3 remainder i s2 hence not sufficient hence A
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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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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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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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12 Aug 2013, 11:31
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 4 + 7n is divid
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11 Nov 2013, 19:02
Bunuel wrote: 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. I am not sure if it has been asked/discussed before, but can we use the following approach: rewrite 4 + 7n as (3+1) + (6n + 1). now if we divide this by 3 we are left with (1) + (2n +1), which is essentially (2n + 2) > 2(n+1)/3 leaves no remainder or in other words 0  using statement 1 information. Please correct me if I this approach is incorrect.
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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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12 Nov 2013, 02:03
vaishnogmat wrote: Bunuel wrote: 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. I am not sure if it has been asked/discussed before, but can we use the following approach: rewrite 4 + 7n as (3+1) + (6n + 1). now if we divide this by 3 we are left with (1) + (2n +1), which is essentially (2n + 2) > 2(n+1)/3 leaves no remainder or in other words 0  using statement 1 information. Please correct me if I this approach is incorrect. 4 + 7n = (3+1) + (6n + n) not (3+1) + (6n + 1). You can solve (1) in another way: \(4 + 7n = 4 + 4n + 3n = 4(n + 1) + 3n\). First statement says that \(n + 1\) is is divisible by 3, thus \(4(n + 1) + 3n = (a \ multiple \ of \ 3) + (a \ multiple \ of \ 3)\). Therefore \(4 + 7n\) yields the remainder of 0, when divided by 3. Hope it helps.
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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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06 Aug 2016, 08:28
I had a less algebraic approach, let me know what you think: \(4+7n\) can be expressed in numbers where n is positive integer: \(4+7(1) = 11\) \(4+7(2) = 18\) you get the idea. Statement 1) \(n+1\) is divisible by 3: So we should experiment with a few n+1's .... if \(n=2\), then \(n+1 = 3\) is divisible by \(3\). \((4+(7*2)) / 3 = 18 / 3 = 6\) with \(0\) remainder. if \(n = 5\), then \(n+1=6\) which is divisible by 3. \((4+7(5))/3 = 39 / 3 = 13\) with \(0\) remainder. So with those 2 I assumed it is sufficient. Statement 2) is clearly insufficient. Greetings!
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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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30 Jan 2017, 22:58
1) n+1 is a multiple of 3. If n=2 (we're allowed to pick 2 since 2+1 is a multiple of 3), then (4+14)/3 = 18/3 = 6rem0 If n=5 (we're allowed to pick 5 since 5+1 is a multiple of 3), then (4+35)/3 = 39/3 = 13rem0 at this point you might already be conviced that you'll always get the same answer, but we could try one more just to be safe: If n=8 (we're allowed to pick 8 since 8+1 is a multiple of 3), then (4+56) = 60/3 = 20rem0 For all 3 plugins we get r=0.. sufficient! 2) n > 20 If n=21, then (4+147)/3 = 151/3 = 50rem1 Insuff. Hence A.
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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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16 Mar 2019, 09:24
Hi, Here are my two cents for this question we are told that n>0 and r is remainder when the expression 4+7n is divided by 3. let the expression be x so we have x= \(\frac{4+7n}{3}\) yields a reminder r where 0\(\leq{r}\)\(<\)3 Statement 1 Says n+1 is divisible by 3 , which means (n+4), (n+7)... are also divisible by 3 Rewriting the expressions x = 6n+n+4 x= \(\frac{6n+n+4}{3}\) we have \(\frac{6n}{3}\) + \(\frac{n+4}{3}\) since both the fractions are divisible by 3 we have remainder is 0 So r= 0 Hence A is sufficient. Statement 2: n>20 we can write expression x as x= \(\frac{4}{3}\) + \(\frac{6n}{3}\)+ \(\frac{n}{3}\) now we see that remainder will be 1+0+ reminder from the part \(\frac{n}{3}\) if n is multiple of 3 reminder will be 1 if n is 1 more than multiple of 3 reminder will be 2 if n is 2 more than multiple of 3 reminder will be 0 Hence B is insufficeint
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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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22 Jun 2019, 10:51
Bunuel is this another valid explanation for S1: 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 If n+1 is divisible by 3, n gives us a remainder of 2 (n is positive integer). Then 7n also gives a remainder of 2. Then 7n +1 is divisible by 3, and also 7n +4 (every three numbers we can find one which is divisible by 3). > A Sufficient.



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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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14 Sep 2019, 06:54
HiteshPunjabi  have you used rebuilding the dividend equation?




Re: If n is a positive integer and r is the remainder when 4 + 7n is divid
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