If n is a positive integer and r is the remainder when 4 + 7n is divid

If n is a positive integer, and r is the remainder when 4 + 7n is divided by 3, what is the value of r?

The condition "r is the remainder when 4 + 7n is divided by 3" can be expressed as \(4 + 7n = 3q + r\), where r is an integer such that \(0 \leq r < 3\). The question asks for the value of r.

(1) n + 1 is divisible by 3

This implies \(n + 1 = 3k\), or \(n = 3k - 1\). Substituting this value of n into the equation, we get \(4 + 7(3k - 1) = 3q + r\). Simplifying, this becomes \(3(7k - 1 - q) = r\). This means that r is a multiple of 3. However, since r is an integer within the range \(0 \leq r < 3\), the only possibility is r = 0. Sufficient.

(2) n > 20.

This is clearly not sufficient. For example, if \(n = 21\), then \(4 + 7n = 151\), resulting in \(r = 1\). However, if \(n = 22\), then \(4 + 7n = 158\), resulting in \(r = 2\).

Answer: A.
_________________

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

Answer is (a)
_________________

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.
_________________

Can you explain how you get to 7n+4=3(2n+1)?

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.

Just trying to split it out into parts divisible by 3
7n becomes 6n+n
4 becomes 3+1
_________________

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

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!

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.
_________________

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.

Info from the stem:
1. N > 0 and integer
2. 7N + 4 when divided by 3 leaves a remainder R

7N + 4 can be re-written as 6N + 3 + (N + 1). 6N and 3 are divisible by 3 always, so we need to find what is the remainder when N+1 is divided by 3.

S1: N+1 is divisible by 3 - implies R=0 (unique answer). Sufficient.
S2: N > 20 - multiple answers are possible. Insufficient.

Hence the correct answer is option A.

Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

I'm thinking of the following approach. Please tell me if it's incorrect.

if r is the reminder when 4+7n is divided by 3 means r = 1(reminder of 4/3) + reminder of 7n/3
Now,
option-1 - if n+1 is divisible by 3, reminder of n when divided by 3 is 2. hence now r = 1 + 2 = 3 which is divisible by 3 and hence r =0. Sufficient.
However,
option-2 - says n > 20. This doesn't mean anything as it can have multiple values which may or may not be divisible by 3. In sufficient.

Answer - A

Good explanation Bunuel
One question not sure how you derive below? Could you elaborate please? Thanks

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

\(4+7(3k-1)=3q+r\);

\(4+21k-7=3q+r\);

\(21k-3=3q+r\);

\(21k-3-3q=r\);

\(3(7k-1-q)=r\).

Hope this helps.
_________________

Ah get it thanks Bunuel , I need to remember to use factoring.
Thanks once again for your great help always !!

Hi Bunuel. Is my method correct? Do we really need to use r in this solution? Thank you!

With statement 1, n+1 = 3K (since n+1 is divisible by 3)
Replacing value of n in the original number 7n+4 = n(3k-1)+4 = 21k - 3 = 3(7k-1), which is divisible by 3. Therefore, remainder is 0.