A digit n > 3 is divisible by 3 but not divisible by 6. Which one of the following is divisible by 4?

Correct Option: 3n + 1

Let us analyze the given condition for the digit $n$.

We are given that $n>3$, and $n$ is a single digit divisible by 3, but not divisible by 6.

The single-digit numbers greater than 3 are 4, 5, 6, 7, 8, and 9.

From these digits, let us find the digits divisible by 3:

- 6 is divisible by 3, but 6 is also divisible by 6 (since 6 is even). Thus, $n$ cannot be 6.
- 9 is divisible by 3 and is NOT divisible by 6 (since 9 is odd). Thus, $n=9$.

Now, let us test $n=9$ in each of the given options to determine which one is divisible by 4:

1. Option 1 (2n):

$2n=2\times 9=18$

18 is not divisible by 4 (18 ÷ 4 = 4 with a remainder of 2).

2. Option 2 (3n):

$3n=3\times 9=27$

27 is not divisible by 4 (27 is odd).

3. Option 3 (2n + 4):

$2n+4=2\left(9\right)+4=18+4=22$

22 is not divisible by 4 (22 ÷ 4 = 5 with a remainder of 2).

4. Option 4 (3n + 1):

$3n+1=3\left(9\right)+1=27+1=28$

28 is completely divisible by 4 ($28÷4=7$).

Therefore, the expression 3n + 1 is divisible by 4.