10.71 Given that $k$ and $l$ are natural numbers. Find the number of the form $36^{k}-5^{l}$ with the smallest absolute value, and prove that the number found is indeed the smallest.
(8th All-Soviet Union Mathematical Olympiad, 1974)
[Solution] Since the unit digit of $36^{k}$ is always 6, and the unit digit of $5^{l}$ is always 5, when $36^{k}>5^{l}$, the unit digit of $\left|36^{k}-5^{l}\right|$ is 1, and when $36^{k}<5^{l}$, the unit digit of $\left|36^{k}-5^{l}\right|$ is 9.
First, consider the case where the unit digit of $\left|36^{k}-5^{l}\right|$ is 1.
If $36^{k}-5^{l}=1$, then
$$
5^{l}=36^{k}-1=\left(6^{k}+1\right)\left(6^{k}-1\right).
$$

Obviously,
$$
5 \nmid 6^{k}+1.
$$

Therefore, $36^{k}-5^{l}=1$ does not hold.
If $36^{k}-5^{l}=11$, this is possible, in which case $k=1, \quad l=2$.
Next, consider the case where the unit digit of $\left|36^{k}-5^{l}\right|$ is 9.
If $5^{l}-36^{k}=9$, then by $319,3 \mid 36^{k}, 31$ it is known that this is impossible.
Therefore, the smallest absolute value of the form $\left|36^{k}-5^{l}\right|$ is
$$
36-5^{2}=11
$$