1. The answers are $a=1098, b=11097$ and $c=111096$.
2. Let $N=99+999+\cdots+\overbrace{99 \ldots 9}^{100}$. This sum has 99 terms. We write each term differently:

$$
99=100-1,999=1000-1 \ldots \text {, and } \overbrace{99 \ldots 9}^{100}=1 \overbrace{00 \ldots 0}^{100}-1
$$

$$
\begin{aligned}
N & =(100-1)+(1000-1)+\cdots+1 \overbrace{00 \ldots 0}^{100}-1 \\
& =\overbrace{11 \ldots 11}^{99} 00-99 \\
& =\overbrace{11 \ldots 1}^{98} 001
\end{aligned}
$$

The digit 1 appears 99 times.

3. We first notice that $M=N / 9$. Therefore, we need to divide this large number by 9.

$$
\begin{aligned}
& \overbrace{111111111}^{9} \div 9=012345679, \text { so } \\
& \overbrace{1111111111111111111 \ldots 1111}^{99} \div 9= \\
& 012345679012345679 \ldots 5679
\end{aligned}
$$

where the sequence 012345679 is repeated 11 times.

$$
\begin{aligned}
M & =(\overbrace{111111111111111111 \ldots 11}^{99} 00-99) \div 9 \\
& =012345679012345679 \ldots 567900-11 \\
& =012345679012345679 \ldots 567889
\end{aligned}
$$

The digit 1 is written 11 times.
1. Calculate the sums: $a=99+999, b=99+999+9999$ and $c=99+999+9999+99999$.
2. Consider the number $N$ defined as the sum:

$$
N=99+999+9999+\cdots+9999999 \ldots 999
$$

The first term of this sum is written with two nines; then we add the numbers written with three, four nines, etc. The last term of the sum is written with one hundred nines. We perform the sum and write $N$ in ordinary decimal notation. How many times does the digit 1 appear in this notation?

3. Same question for the number $M=11+111+1111+\cdots+1111111 \ldots 111$ where the last term of the sum is written with one hundred ones.