6.8 Let $a=19^{91}, b=(999991)^{19}$, then $a-b$ is
(A) a number not less than 1.
(B) a number with absolute value greater than 0 and less than 1.
(C) a number not greater than -1.
(D) 0.
(China Beijing Junior High School Mathematics Competition, 1991)
[Solution 1] From $b=(999991)^{19}19^{76}\left(19^{15}-2^{57}\right)>19^{76}\left(16^{15}-2^{57}\right)$
$$
\begin{array}{l}
=19^{76}\left(2^{60}-2^{57}\right)=19^{76} \cdot 2^{57} \cdot\left(2^{3}-1\right) \\
=19^{76} \cdot 2^{57} \cdot 7
\end{array}
$$

Therefore, the answer is $(A)$.
[Solution 2]
$$
\begin{array}{l}
b=(999991)^{19}19^{76} \cdot 18^{15}=19^{76} \cdot 2^{15} \cdot 9^{15} \\
>19^{76} \cdot 2^{15} \cdot 8^{15}=19^{76} \cdot 2^{60},
\end{array}
$$
$\therefore a>b$, and since $a, b$ are integers, then $|a-b| \geqslant 1$.
Therefore, the answer is $(A)$.