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

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Re: GMAT PREP (DS) [#permalink]

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New post 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 \(n=3k-1\) --> \(4+7(3k-1)=3q+r\) --> \(3(7k-1-q)=r\) --> 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.


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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New post 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 \(n=3k-1\) --> \(4+7(3k-1)=3q+r\) --> \(3(7k-1-q)=r\) --> 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.



I am not sure if it has been asked/discussed before, but can we use the following approach:

re-write 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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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 \(n=3k-1\) --> \(4+7(3k-1)=3q+r\) --> \(3(7k-1-q)=r\) --> 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.



I am not sure if it has been asked/discussed before, but can we use the following approach:

re-write 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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Re: If n is a positive integer, and r is the remainder when [#permalink]

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New post 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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New post 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.

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

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New post 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 \(n=3k-1\) --> \(4+7(3k-1)=3q+r\) --> \(3(7k-1-q)=r\) --> 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.


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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New post 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 \(n=3k-1\) --> \(4+7(3k-1)=3q+r\) --> \(3(7k-1-q)=r\) --> 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.


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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New post 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 plug-ins 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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New post 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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New post 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.

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Re: If n is a positive integer, and r is the remainder when   [#permalink] 22 Sep 2017, 10:46

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