99. When dividing a five-digit number, consisting of identical digits, by a four-digit number, also consisting of identical digits, the quotient was 16 and some remainder. After discarding one digit from both the dividend and the divisor, the quotient remained unchanged, but the remainder decreased by 2000. Find these numbers.
Solution. Let the required numbers be $\overline{a a a a a}$ and $\overline{b b b b}$. According to the problem,

$$
\begin{aligned}
& \overline{a a a a a}=\overline{b b b b} \cdot 16+r \\
& \overline{a a a a}=\overline{b b b} \cdot 16+(r-2000)
\end{aligned}
$$

where $r$ is the remainder of the division of $\bar{a} \overline{a a a a}$ by $\overline{b b b b}$. After subtracting the second equation from the first, we get:

$$
a \cdot 10^{4}=b \cdot 10^{3} \cdot 16+2000, \text { i.e. } 5 a=8 b+1
$$

142

Since $a$ and $b$ are single-digit numbers and the right-hand side of the last equation is an odd number, $a$ must be an odd number. For $a=1,3,7,9$, we get that $b$ is not an integer (since $b$ is a digit), but for $a=5$ we get $b=3$, so the required numbers are 55555 and 3333.

Answer: 55555 and 3333.