190. Among the numbers of the form $36^{k}-5^{l}$, where $k$ and $l$ are natural numbers, find the smallest in absolute value. Prove that the found number is indeed the smallest.
190. Answer: $11=36-5^{2}$.

The last digit of the number $36^{h}=6^{2 k}$ is 6, the last digit of the number $5^{t}$ is 5. Therefore, the number $\left|6^{2 k}-5^{t}\right|$ ends in either 1 (if $6^{2 k}>5^{t}$) or 9 (if $6^{2 k}<5^{t}$).

The equation $6^{2 k}-5^{t}=1$ is impossible, because then it would be $5^{l}=\left(6^{k}-1\right)\left(6^{k}+1\right)$, and the number $6^{k}+1$ is not divisible by 5. For $k=1$ and $l=2$, we get $36^{k}-5^{t}=11$.

The equation $5^{t}-6^{2 k}=9$ is also impossible, because $5^{t}$ for a natural $l$ is not divisible by 3.