14) Let $m=999 \ldots 99$ be the number formed by 77 digits all equal to 9 and let $n=777 \ldots 77$ be the number formed by 99 digits all equal to 7. The number of digits of $m \cdot n$ is
(A) 175
(B) 176
(C) 177
(D) 7692
(E) 7693.
14) The answer is $\mathbf{( B )}$.

We have $m \cdot n=(m+1) \cdot n-n$. Since the number $(m+1) \cdot n=10^{77} \cdot n$ is obtained by adding 77 zeros to $n$, it has $99+77=176$ digits, the first of which is 7. By subtracting $n$, which has fewer digits, the first digit can decrease by at most one, and thus the number of digits does not change.

SECOND SOLUTION

Indeed, since $m9 \cdot 10^{76}, n>7 \cdot 10^{98}$, it follows that $m \cdot n>63 \cdot 10^{175}$, i.e., $m \cdot n$ has at least 176 digits.