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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.

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.

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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It's not just in some of us; it's in everyone. And as we let our own light shine, we unconsciously give other people permission to do the same.

As we are liberated from our own fear, our presence automatically liberates others.

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.

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

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26 Jan 2016, 19:35

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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 \(n=3k-1\) --> \(4+7(3k-1)=3q+r\) --> \(3(7k-1-q)=r\) --> 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?

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

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

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).

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