6.99 $k$, $l$ are natural numbers, find the number with the smallest absolute value in the form of $36^{k}-5^{l}$, and prove that the number found is indeed the smallest.

In the form of $36^{k}-5^{l}$, 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 $5$. Therefore, 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 $\left|36^{k}-5^{l}\right|=1$, i.e., $\square$
$$36^{k}-5^{l}=1$$

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 $\left|36^{k}-5^{l}\right|=9$,
i.e., $5^{l}-36^{k}=9$, $\square$
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 $\left|36^{k}-5^{l}\right|=11$,
i.e., $36^{k}-5^{l}=11$.
Clearly, 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.