matrices with huge numbers as components 
If $A=\begin{bmatrix}
 10^{30}+5& 10^{20}+4 &10^{20}+6 \\ 
10^{4}+2 & 10^{8}+7 &10^{10}+2n \\ 
 10^{4}+8&10^{6}+4  &10^{15}+9 
\end{bmatrix}$ for all $n\in \mathbb{N},$ Then
$(a)\;\;A$ is invertible for all $n\in \mathbb{N}$
$(b)\;\;A$ is not  invertible for all $n\in \mathbb{N}$
$(c)\;\;A$ may or may not be invertible depending on the values of $n\in \mathbb{N}$
$(d)\;$ Data Insufficient

What I try
If $A$ is invertiable, then $\det(A)\neq 0$
$$A=\begin{vmatrix}
 10^{30}+5& 10^{20}+4 &10^{20}+6 \\ 
10^{4}+2 & 10^{8}+7 &10^{10}+2n \\ 
 10^{4}+8&10^{6}+4  &10^{15}+9 
\end{vmatrix}$$
Expanding along $1^\mathrm{st}$ row
$\displaystyle A=\bigg(10^{30}+5\bigg)\bigg[\bigg(10^8+7\bigg)\bigg(10^{15}+9\bigg)-\bigg(10^6+4\bigg)\bigg(10^{10}+2n\bigg)\bigg]-\bigg(10^{20}+4\bigg)\bigg[\bigg(10)^4+2\bigg)\bigg(10^{15}+9\bigg)-\bigg(10^{10}+2n\bigg)\bigg(10^8+4\bigg)\bigg]+\bigg(10^{20}+6\bigg)\bigg[\bigg(10^4+2\bigg)\bigg(10^6+4\bigg)-\bigg(10^8+7\bigg)\bigg(10^4+8\bigg)\bigg]$
How do I simplify such a huge calculation?
Please help me. 
 A: Using the modulo 2 hint,
the principal diagonal is odd and since all the other elements are even, every other component of the determinant, including the 2n term, is even, so the determinant is odd and therefore not zero.
(Added later)
Note that the applies to any matrix with all the diagonal elements odd and the off diagonal elements even.
A: Following @Dirk's hint. we can actually take each component modulo $2$, since the determinant is a polynomial function of $A$'s entries with integer coefficients. So the determinant has the same parity as $\det I_3=1$. For each $n$, $\det A$ is odd so isn't $0$, making the answer (a).
A: Write for $A$
$$
A=\left[
\begin{array}{cc}
 o_1\textrm{  }e_1\textrm{  }e_2\\
 e_3\textrm{  }o_2\textrm{  }e_4\\
 e_5\textrm{  }e_6\textrm{  }o_3
\end{array}\right],
$$
where $o_i=$odd, $i=1,2,3$ and $e_i=$even, $i=1,\ldots,6$. Then
$$
\textrm{det}(A)=\left|A\right|=e_1e_4e_5+e_2e_3e_6-e_4e_6o_1-e_2e_5o_2-e_1e_3o_3+o_1o_2o_3=\textrm{odd.}
$$
Hence $|A|\neq 0$ and thus $A$ is invertable
