23 Let $k$ and $m$ be positive integers. Find the minimum possible value of $\left|36^{k}-5^{m}\right|$.
23. Notice that
$$36^{k}-5^{m} \equiv \pm 1(\bmod 6)$$

and
$$36^{k}-5^{m} \equiv 1(\bmod 5)$$

Therefore, the positive integers that $36^{k}-5^{m}$ can take, in ascending order, are $1, 11, \cdots$.
If $36^{k}-5^{m}=1$, then $k>1$, so taking both sides modulo 8, we require
$$5^{m} \equiv-1(\bmod 8)$$

However, $5^{m} \equiv 1$ or $5(\bmod 8)$, which is a contradiction.
Additionally, when $k=1, m=2$, we have $36^{k}-5^{m}=11$. Therefore, the smallest positive integer that $36^{k}-5^{m}$ can take is 11.

Now consider the smallest positive integer that $5^{m}-36^{k}$ can take. From
$$\begin{array}{l}
5^{m}-36^{k} \equiv 4(\bmod 5) \\
5^{m}-36^{k} \equiv \pm 1(\bmod 6)
\end{array}$$

we know that
$$5^{m}-36^{k} \geqslant 19$$

In summary, the smallest possible value of $\left|36^{k}-5^{m}\right|$ is 11.