If n is an integer, and r is the remainder when 4n+7 is

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

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.

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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)