How to tell if system of congruences where each base is a power of prime $p$ has a solution $p$ is a prime number.
How to tell if a system of congruences:
\begin{align} 
x &\equiv  a_1 \pmod{p^{i_1}} \\
x&\equiv  a_2 \pmod{p^{i_2}} \\
&\dots\\
x &\equiv  a_n \pmod{p^{i_n}}
\end{align} 
Has a solution.
How would you find the solution?
I feel like it has something to do with the chinese remainder theorem but the mod bases are definitely not pairwise relatively prime
 A: If I understand your question correctly, you have
\begin{align*}
x & \equiv a_1 \pmod{p^{e_1}}\\
x & \equiv a_2 \pmod{p^{e_2}}\\
\vdots & \equiv \vdots\\
x & \equiv a_k \pmod{p^{e_k}}
\end{align*}
Without the loss of generality, let us assume $e_1 \leq e_2 \leq \dotsb \leq e_k$.  If the last congruence $x \equiv a_k \pmod{p^{e_k}}$ has a solution, then we should have $x \equiv a_k \pmod{p^j}$ for all $j \leq e_k$. In other words, all the congruences with smaller powers should also give the same residue. So you need $a_1 \equiv a_2 \equiv \dotsb$ for the solution to exist.
A: The general condition for a system of congruences with non-necessarily coprime moduli
$$x\equiv a_i\pmod{m_i}\qquad (1\le i\le r)$$
is that
$$\forall (i,j), 1\le i,j\le r,\enspace a_i\equiv a_j\mod \gcd(m_i,m_j).$$
A: In general,  $x \equiv a \pmod{p^\alpha}$ and  $x \equiv b \pmod{p^\beta}$ if and only if $a \equiv b \pmod{p^\beta}$.
For example, $x \equiv a \pmod{13^5}$ and  $x \equiv b \pmod{13^3}$ if and only if $a \equiv b \pmod{13^3}$
