$6 \cdot 99 k 、 l$ are natural numbers, in the numbers of the form $36^{k}-5^{l}$ find the number with the smallest absolute value, and prove that the number found is indeed the smallest.

In the numbers of the form $36^{k}-5^{l}$, find the number with the smallest absolute value, and prove that the number found is indeed the smallest.
[Solution] The last digit of $36^{k}$ is $6$, and the last digit of $5^{l}$ is $5$, so when $36^{k}>5^{l}$, the last digit of $\left|36^{k}-5^{l}\right|$ is 1; when $36^{k}<5^{l}$, the last digit of $\left|36^{k}-5^{l}\right|$ is 9.
If $\quad\left|36^{k}-5^{l}\right|=1$,

that is
$$
36^{k}-5^{l}=1,
$$

then we have $\quad 5^{l}=\left(6^{k}-1\right)\left(6^{k}+1\right)$.
But $6^{k}+1$ is not a multiple of 5, so this situation cannot occur.
If $\quad\left|36^{k}-5^{l}\right|=9$,

that is
$$
5^{l}-36^{k}=9 \text {, }
$$

then we get $\quad 5^{l}=36^{k}+9$.
But the left side of the above equation is not a multiple of 3, while the right side is a multiple of 3, which is a contradiction. Therefore, this situation cannot occur either.
If $\quad\left|36^{k}-5^{l}\right|=11$,

that is $36^{k}-5^{l}=11$.
Obviously, when $k=1, l=2$, the above equation holds.
Therefore, among numbers of the form $36^{k}-5^{l}$, the number with the smallest absolute value is 11.