1. Find the smallest positive integer $k$, such that there exist positive integers $m, n$, satisfying $k=19^{n}-5^{m}$.
1. Solution: $k_{\text {min }}=14$ Hint: On one hand, take $m=n=1$, then $k=14$. On the other hand, let $k=19^{n}-5^{m}$.

Modulo 2 gives: $k \equiv 0(\bmod 2)$;
Modulo 3 gives: $k \equiv 1^{n}-(-1)^{m} \equiv 0$ or $2(\bmod 3)$;
Modulo 5 gives: $k \equiv(-1)^{n}(\bmod 5)$.
Based on the above conditions, we can determine that $k \neq 1,2,3,4,5,7,8,9,10,11,12,13$.
If $k=6$, then modulo 19 gives $5^{m} \equiv 13(\bmod 19)$, which is impossible (by checking $5^{1}, 5^{2}, \cdots, 5^{9}$, we find they are congruent to $5,6,11,17,9,7,16,4,1(\bmod 19))$, so $k \geqslant 14$.
Combining both aspects, the minimum value of $k$ is 14.