, 2016

Solution:
Note that
$$2^{10 n} = 1024^{n} = 1.024^{n} \times 10^{3 n}$$
So $2^{10 n}$ has roughly $3 n + 1$ digits for relatively small $n$'s. (Actually we have that for $0 < x < 1$,
$$(1 + x)^{2} = 1 + 2x + x^{2} < 1 + 3x$$
Therefore, $1.024^{2} < 1.03^{2} < 1.09$, $1.09^{2} < 1.27$, $1.27^{2} < 1.81 < 2$, and $2^{2} = 4$, so $1.024^{16} < 4$. Thus the conclusion holds for $n \leq 16$.)

For any positive integer $n \leq 16$,
$$A = \sum_{i=1}^{n} 2^{10 i} \times 10^{\sum_{j=i+1}^{n} (3j + 1)}$$
Let
$$A_{i} = 2^{10 i} \times 10^{\sum_{j=i+1}^{n} (3j + 1)}$$
for $1 \leq i \leq n$, then we know that
$$A - 2^{10 n} = \sum_{i=1}^{n-1} A_{i}$$
and
$$A_{i} = 2^{10 i + \sum_{j=i+1}^{n} (3j + 1)} \times 5^{\sum_{j=i+1}^{n} (3j + 1)} = 2^{u_{i}} \times 5^{v_{i}}$$
where $u_{i} = 10 i + \sum_{j=i+1}^{n} (3j + 1)$, $v_{i} = \sum_{j=i+1}^{n} (3j + 1)$. We have that
$$u_{i} - u_{i-1} = 10 - (3i + 1) = 3(3 - i)$$
Thus, for $1 \leq i \leq n-1$, $u_{i}$ is minimized when $i = 1$ or $i = n-1$, with $u_{1} = \frac{3 n^{2} + 5 n + 12}{2}$ and $u_{n-1} = 13 n - 9$. When $n = 5$,
$$A - 2^{10 n} = A_{1} + A_{2} + A_{3} + A_{4} = 2^{10} \times 10^{46} + 2^{20} \times 10^{39} + 2^{30} \times 10^{29} + 2^{40} \times 10^{16}$$
is at most divisible by $2^{57}$ instead of $2^{170}$. For all other $n$'s, we have that $u_{1} \neq u_{n-1}$, so we should have that both $170 \leq u_{1}$ and $170 \leq u_{n-1}$. Therefore, since $170 \leq u_{n-1}$, we have that $14 \leq n$. We can see that $u_{1} > 170$ and $14 < 16$ in this case. Therefore, the minimum of $n$ is $14$.