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Re: GMAT PREP (DS) [#permalink]
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14 Sep 2013, 02:29
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. another way of seeing it 4+7n = 3+6n +n+1; 3+6n is divisible by 3; n+1 is divisible by 3; so 4+7n is divisible by three.



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Re: GMAT PREP (DS) [#permalink]
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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: GMAT PREP (DS) [#permalink]
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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 [#permalink]
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09 Apr 2016, 14:36
my approach 4+4n+3n 4+4n is divisible by 3 3n is divisible by 3 so r=0
2 alone can yield multiple answers. so no.



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Re: If n is a positive integer, and r is the remainder when [#permalink]
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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 [#permalink]
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11 Oct 2016, 07:48
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. is there any other method to solve this problem?? can we solve it by plugging in numbers?



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Re: If n is a positive integer, and r is the remainder when [#permalink]
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11 Oct 2016, 09:21
nishantdoshi 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. is there any other method to solve this problem?? can we solve it by plugging in numbers? I solved it by plugging in the numbers. Statement 2 is clearly insufficient, so I am not discussing more on that. Statement 1 : n+1 is divisible by 3. It means n is not going to be a multiple of 3. Also, n will be 1 less than the multiple of 3. Try taking the values of n as 2,5,8,20,23. In all the cases, we will get the original expression a multiple of 3. or remainder, r = 0; hence A.
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Re: If n is a positive integer, and r is the remainder when [#permalink]
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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 [#permalink]
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15 Jul 2017, 14:33
"If n is a positive integer and r is the remainder when 4+7n is divided by 3, what is the value of r?"
In other words... find "r" when
\(r ≡ 7n + 4 (mod 3)\)
Which can be simplified to:
\(r ≡ 7n + 1 (mod 3)\)
Right away, we see a pattern: When \(n = 1, r =2\). When \(n = 2, r =0\). When \(n = 3, r = 1.\) ... And the pattern repeats over and over.
Statement A.)
"n+1 is divisible by 3"
In other words: \(n + 1 ≡ 0 (mod 3)\)
Which is equivalent to... \(n ≡ 2 (mod 3)\)
So since we now know "n," we can plug in its value of "2" for our original congruence...
\(r ≡ 7n + 1 (mod3)\)
\(r ≡ 7*2 + 1 (mod3)\) \(r ≡ 15 (mod3)\) \(r ≡ 0(mod3)\)
...and we find that the remainder is ZERO.
SUFFICIENT.
Statement B.)
"n>20"
Remembering our pattern from earlier (When \(n = 1, r =2\). When \(n = 2, r =0\). When \(n = 3, r = 1\)......etc)
Statement B is obviously insufficient. At values of "n" greater than 20, the value of "r" continues to cycle \(0,1,2,0,1,2,0,1,2.... etc\), so we have no idea what "r" is equal to without knowing a specific value for "n" (in mod 3).
Answer is A.



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Re: If n is a positive integer, and r is the remainder when [#permalink]
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22 Sep 2017, 10:46
jordanhendrix 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 or, given that n+1 is divisible by 3 7n+4 = 7(n+1)3 and because both 7(n+1) and 3 are divisible by 3, the remainder r = 0.




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