The sequence $a_0,a_1,a_2,\cdots$ is a strictly increasing arithmetic sequence of positive integers such that \[2^{a_7}=2^{27} \cdot a_7.\] What is the minimum possible value of $a_2$?
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 22$
We can rewrite the given equation as $2^{a_7-27}=a_7$. Hence, $a_7$ must be a power of $2$ and larger than $27$. The first power of 2 that is larger than $27$, namely $32$, does satisfy the equation: $2^{32 - 27} = 2^5 = 32$. In fact, this is the only solution; $2^{a_7-27}$ is exponential whereas $a_7$ is linear, so their graphs will not intersect again.
Now, let the common difference in the sequence be $d$. Hence, $a_0 = 32 - 7d$ and $a_2 = 32 - 5d$. To minimize $a_2$, we maxmimize $d$. Since the sequence contains only positive integers, $32 - 7d > 0$ and hence $d \leq 4$. When $d = 4$, $a_2 = \boxed{\textbf{(B)}\ 12}$.
~[Bxiao31415](https://artofproblemsolving.com/wiki/index.php/User:Bxiao31415)