<s>[INST] Least positive integer $n$ such that the digit string of $2^n$ ends on the digit string of $n$ 
What is the least positive integer $n$ such that the digit string of $2^n$ ends on the digit string of $n$:
  $$ (2^n)_{10} = d_m \, d_{m-1} \cdots
 d_{q+1} \, (n)_{10} \\ (n)_{10} = d'_{q} \cdots d_1' \\ d_i, d'_j \in
 \{0, \ldots, 9 \} $$

As in $2^3$ would somehow end in 3, or $2^5$ would end in 5.
Frankly I don't even know where to start. Thanks in advance.
 [/INST] Here is a less brute-force method:
In order for the digit string of $2^n$ to end with the digit string of $n$, it is necessary (but not sufficient) that $2^n$ and $n$ have the same last digit, i.e. $2^n \equiv n \pmod{10}$. 
It is easy to check that $\begin{cases}2^n \equiv 2\pmod{10} & \text{if} \ n\equiv 1 \pmod{4} \\ 2^n \equiv 4\pmod{10} & \text{if} \ n\equiv 2 \pmod{4} \\ 2^n \equiv 8\pmod{10} & \text{if} \ n\equiv 3 \pmod{4} \\ 2^n \equiv 6\pmod{10} & \text{if} \ n\equiv 0 \pmod{4}\end{cases}$. 
Since $2^n$ is even, $n$ must also be even. So we have the following 2 possibilities:
$n \equiv 2\pmod{4}$ AND $n \equiv 2^n \equiv 4\pmod{10}$, i.e. $n \equiv 14 \pmod{20}$
$n \equiv 0\pmod{4}$ AND $n \equiv 2^n \equiv 6\pmod{10}$, i.e. $n \equiv 16 \pmod{20}$. 
So we only need to check $n \equiv 14 \ \text{or} \ 16 \pmod{20}$. 
Since $2^{14} = 16384$, $2^{16} = 65536$, $2^{34} = 17179869184$, and $2^{36} = 68719476736$, the smallest positive integer $n$ such that the digit string of $2^n$ ends with the digit string of $n$ is $n = 36$.
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