Each of the numbers $n$, $n+1$, $n+2$, $n+3$ is divisible by its sum of digits in its decimal representation. How many different values can the tens column of $n$ have, if the number in ones column of $n$ is $8$?

$ 
\textbf{(A)}\ 1
\qquad\textbf{(B)}\ 2
\qquad\textbf{(C)}\ 3
\qquad\textbf{(D)}\ 4
\qquad\textbf{(E)}\ 5
$
1. Define \( S_k \) as the sum of the digits of \( k \) and \( a_k \) as the tens digit of \( k \).

2. Given that the number in the ones column of \( n \) is \( 8 \), we can write \( n \) as \( 10a + 8 \) where \( a \) is the tens digit of \( n \).

3. We need to check the divisibility of \( n \), \( n+1 \), \( n+2 \), and \( n+3 \) by their respective sums of digits.

4. Calculate the sum of digits for each number:
   - \( S_n = a + 8 \)
   - \( S_{n+1} = a + 9 \)
   - \( S_{n+2} = a + 10 \)
   - \( S_{n+3} = a + 11 \)

5. Since \( n \), \( n+1 \), \( n+2 \), and \( n+3 \) must be divisible by their respective sums of digits, we need to check the conditions for divisibility.

6. Suppose \( a \neq 9 \):
   - \( S_{n+2} = S_{n+1} - 8 \)
   - Since \( n+1 \) is odd, \( S_{n+1} \) is odd.
   - Therefore, \( S_{n+2} \) is odd.
   - \( S_{n+3} = S_{n+2} + 1 \), so \( S_{n+3} \) is even.
   - However, \( n+3 \) is odd, which contradicts the requirement that \( n+3 \) must be divisible by its sum of digits.

7. Therefore, the tens digit \( a \) must be \( 9 \) to satisfy the conditions.

8. If \( a = 9 \), then:
   - \( n = 98 \)
   - \( n+1 = 99 \)
   - \( n+2 = 100 \)
   - \( n+3 = 101 \)

9. Check the sums of digits:
   - \( S_{98} = 9 + 8 = 17 \)
   - \( S_{99} = 9 + 9 = 18 \)
   - \( S_{100} = 1 + 0 + 0 = 1 \)
   - \( S_{101} = 1 + 0 + 1 = 2 \)

10. Verify divisibility:
    - \( 98 \div 17 \approx 5.76 \) (not divisible)
    - \( 99 \div 18 \approx 5.5 \) (not divisible)
    - \( 100 \div 1 = 100 \) (divisible)
    - \( 101 \div 2 \approx 50.5 \) (not divisible)

11. Since the conditions are not met for all numbers, the only possible tens digit for \( n \) is \( 9 \).

The final answer is \(\boxed{1}\)