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Director  Joined: 03 Sep 2006
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If n is a positive integer and r is the remainder when 4 + 7n is divid  [#permalink]

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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.

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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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.

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

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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

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

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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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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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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Why am i getting negative Reminder here?

Given: 4 + 7n = 3q + r

If we simplify this further,

7n = 3q + r -4
n = (3q/7) + (r-4)/7
n + 1 = (3q/7) + (r-4)/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., (r-4/7) + 1 = 0 ==> r = -3

Cheers!
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Re: If n is a positive integer and r is the remainder when 4 + 7n is divid  [#permalink]

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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  [#permalink]

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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

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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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Bunuel wrote:
LM 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?

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.

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: If n is a positive integer and r is the remainder when 4 + 7n is divid  [#permalink]

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vaishnogmat wrote:
Bunuel wrote:
LM 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?

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.

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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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 4 + 7n is divid  [#permalink]

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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 4 + 7n is divid  [#permalink]

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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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GMAT 1: 690 Q50 V34 Re: If n is a positive integer and r is the remainder when 4 + 7n is divid  [#permalink]

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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  [#permalink]

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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   [#permalink] 14 Sep 2019, 06:54
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