Among the numbers of the form $36^{k}-5^{l}$ (where $k, l$ are positive integers), which one has the smallest absolute value?
We will show that this number is 11. Indeed, 11 has this form: $11=36^{1}-5^{2}$. We will prove that any other number of the form $36^{k}-5^{l}$ has an absolute value greater than 11.

Let's examine what remainder a number of the form $36^{k}-5^{l}$ can give when divided by 20!

The powers of 36 always give a remainder of 16: $36 \equiv 16(\bmod 20)$ and if $36^{n} \equiv 16(\bmod 20)$, then

$$
36^{n+1}=36 \cdot 36^{n} \equiv 36 \cdot 16=576 \equiv 16(\bmod 20)
$$

Similarly, it can be shown that $5^{l}$ always gives a remainder of 5: $5^{l} \equiv 5(\bmod 20)$ and if $5^{n} \equiv 5(\bmod 20)$, then $5^{n+1}=5 \cdot 5^{n} \equiv$ $5 \cdot 5=25 \equiv 5(\bmod 20)$. This means that $36^{k}-5^{l}$ always gives a remainder of $16-5=11$ when divided by 20.

Among the numbers that give a remainder of 11 when divided by 20, all (except for -9) have an absolute value greater than 11.

However, $36^{k}-5^{l}$ can never be -9, since $36^{k}$ and 9 are divisible by 3, while $5^{l}$ is not.

Therefore, the number of the form $36^{k}-5^{l}$ with the smallest absolute value is 11.