19. Find the last two digits of the number $99^{99}-51^{51}$.
19. Let's determine the last digits of the powers $99^{99}$ and $5151.99^{1}-99$, $99^{2}-01,99^{3}-99,99^{4}-01$ and so on. $99^{99} - 99$, i.e., $99^{99}=100 M+99$.

Similarly, we establish that $51^{51}=100 N+51$, then $99^{99}-5151=$ $=100(M-N)+99-51=100(M-N)+48$.

Answer: the last two digits are 4 and 8.