1. General Form of $\(n\)$ :

$$
\(n=60 q+9\)
$$


This means $n$ is always 9 more than a multiple of 60 .
2. Check Divisibility by Each Option:

Option A: 3
- 60 is divisible by $\(3(60 \div 3=20)\)$.
- 9 is divisible by $\(3(9 \div 3=3)\)$.
- Thus, $\(n=60 q+9\)$ is always divisible by 3 .
- Conclusion: 3 can divide $n$.

Option B: 6
- For $n$ to be divisible by 6 , it must be divisible by both 2 and 3 .
- From Option A, $n$ is divisible by 3 .
- Now check divisibility by 2 (i.e., is $n$ even?):
- $60 q$ is always even (divisible by 2 ).
- 9 is odd.
- Even + Odd = Odd, so $n$ is always odd.
- Since $n$ is odd, it cannot be divisible by 2 , and thus cannot be divisible by 6 .
- Conclusion: 6 cannot divide $n$.

Option C: 9
- Check if $\(n=60 q+9\)$ is divisible by 9 :
- $\(60 \div 9=6 . \overline{6}\)$ (not divisible).
- However, if $q$ is a multiple of 3 (e.g., $\(q=3 k\)$ ):

$$
\(n=60(3 k)+9=180 k+9=9(20 k+1)\)
$$


This is clearly divisible by 9 .
- Example:
- If $\(q=0, n=9\)$ (divisible by 9 ).
- If $\(q=3, n=189(189 \div 9=21)\)$.
- Conclusion: 9 can divide $n$ (for certain values of $q$ ).

Option D: 12
- For $n$ to be divisible by $12,60 q+9$ must be divisible by 12 .
- 60 is divisible by $\(12(60 \div 12=5)\)$, so $60 q$ is divisible by 12 .
- Thus, $n=60 q+9$ is divisible by 12 only if 9 is divisible by 12 .
- But $\(9 \div 12=0.75\)$ (not an integer).
- No value of $q$ makes $n$ divisible by 12 .
- Conclusion: 12 cannot divide $n$.

Final Answer:
Both $\(\mathbf{B}(6)\)$ and $\(\mathbf{D}(12)\)$ cannot be divisors of $n$.