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

1) (n+1) is divisible by 3
2) n>20

I am having a hard time even starting this one. Please help! Thanks

before we start this discussion, let me remind you of the following important truth:
TAKEAWAY:
WHEN YOU DEAL WITH REMAINDERS, PATTERNS TEND TO EMERGE QUICKLY AND EASILY.

consequence:

if you don't understand the theory in a remainder problem, then just start plugging in numbers and see what sort of patterns emerge.

--

statement 2 is probably easier to start with, since it doesn't have any glitz, glitter, or randomness; it's just a straight inequality. n is greater than 20.
there's not much to work with here, theory-wise, so let's just start plugging in some numbers.
n = 21 --> 4n + 7 = 91 --> remainder = 1 upon division by 3
n = 22 --> 4n + 7 = 95 --> remainder = 2 upon division by 3
insufficient.

--

statement 1:

easier method: JUST PLUG IN NUMBERS
it's not hard to generate plug-ins for this problem: just pick different multiples of 3 to stand in for (n + 1).
n + 1 = 3 --> n = 2 --> 4n + 7 = 15 --> remainder = 0 upon division by 3
n + 1 = 6 --> n = 5 --> 4n + 7 = 27 --> remainder = 0 upon division by 3
n + 1 = 9 --> n = 8 --> 4n + 7 = 39 --> remainder = 0 upon division by 3
n + 1 = 12 --> n = 11 --> 4n + 7 = 51 --> remainder = 0 upon division by 3
there's a clear pattern here: the remainder is always 0.
sufficient.

theory method #1:
you know that n + 1 is a multilple of 3. therefore, you can write n + 1 = 3k, where k is an integer.
subtract to isolate n --> n = 3k - 1.
therefore,
4n + 7
= 4(3k - 1) + 7
= 12k - 4 + 7
= 12k + 3
= 3(4k + 1)
= 3(integer)
therefore, (4n + 7) is a multiple of three. this means it will always yield a remainder of 0 upon division by three.

theory method #2:
instead of isolating n, factor as many (n + 1)'s as possible out of the given quantity.
4n + 7
= (4n + 4) + 3
= 4(n + 1) + 3
= 4(multiple of 3) + 3
= sum of 2 multiples of 3, since 4(multiple of 3) and 3 itself are both multiples of 3
= another multiple of 3
therefore, (4n + 7) is a multiple of three. this means it will always yield a remainder of 0 upon division by three.

ans = (a)

Wow! Thanks Ron!

I believe the question from GMAT Prep is:
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

if the problem features 7n + 4 instead of 4n + 7, then it can be solved with exactly the same methods contained in this thread.
in fact, the answer will still be the same (still (a)).

plugging no is easier here

saves time

Ron, you mean to say in Both the cases OA is A

remainder is always Zero

_________________
Many of the great achievements of the world were accomplished by tired and discouraged men who kept on working.

yes.

i've edited the original posts. thanks.