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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. _________________

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?

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 _________________

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.

Re: Remainder Problem [#permalink]
30 Sep 2010, 21:51

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: Remainder Problem [#permalink]
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 .... _________________

If n is a positive integer, and r is the remainder when 4 + 7n i [#permalink]
26 Oct 2012, 04:47

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.

Re: If n is a positive integer, and r is the remainder when [#permalink]
09 Aug 2013, 08:17

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! _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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

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.

gmatclubot

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

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