Three, (50 points) Write down all positive integers from 1 to 10000 from left to right, then remove those numbers that are divisible by 5 or 7, and concatenate the remaining numbers to form a new number. Try to find:
(1) The number of digits in the new number;
(2) The remainder when the new number is divided by 11.
Three, (1) Obviously, the remainder of the remaining numbers when divided by 35 is
$$
\begin{array}{l}
1,2,3,4,6,8,9,11,12,13,16,17,18, \\
19,22,23,24,26,27,29,31,32,33,34,
\end{array}
$$

denoted as $a_{i}(i=1,2, \cdots, 24)$ in sequence.
Among the numbers from $1 \sim 9$, 7 numbers remain, denoted as $b_{i}$ $(i=1,2, \cdots, 7)$ in sequence.

Among the numbers from $10 \sim 99$, since $99=35 \times 2+29$, there are $24 \times 2+20-7=61$ remaining numbers, denoted as $b_{i}(i=$ $8,9, \cdots, 68)$ in sequence.

Among the numbers from $100 \sim 999$, since $999=35 \times 28+19$, there are $24 \times 28+14-68=618$ remaining numbers, denoted as $b_{i}(i=69,70, \cdots, 686)$ in sequence.
Among the numbers from $1000 \sim 9999$, since
$$
9999=35 \times 285+24 \text {, }
$$

there are $24 \times 285+17-686=6171$ remaining numbers, denoted as $b_{i}(i=687,688, \cdots, 6857)$ in sequence.
In summary, the number of digits in the new number is
$$
7+61 \times 2+618 \times 3+6171 \times 4=26667 \text {. }
$$
(2) First, there are two conclusions:
(i) If $c_{i}(i=1,2, \cdots, k)$ are all $n$-digit positive integers, then
$$
\overline{c_{1} c_{2} \cdots c_{k}} \equiv \sum_{i=1}^{k} c_{i}(-1)^{(k-i) n}(\bmod 11) .
$$
(ii) The sum of any 11 consecutive terms in an arithmetic sequence composed of positive integers is divisible by 11.
Obviously, $b_{24 k+j}=35 k+a_{j}(j=1,2, \cdots, 24)$,
$$
\begin{array}{l}
\sum_{j=1}^{24} b_{24 k+j}=35 k \times 24+\sum_{j=1}^{24} a_{j} \\
\equiv 4 k+2(\bmod 11), \\
\sum_{j=1}^{24}(-1)^{j} b_{24 k+j}=\sum_{j=1}^{24}(-1)^{j} a_{j} \\
\equiv 5(\bmod 11) .
\end{array}
$$

Let the new number be $A=\overline{A_{1} A_{2} A_{3} A_{4}}$, where,
$$
\begin{array}{l}
A_{1}=\overline{1234689} \equiv 1-2+3-4+6-8+9 \\
\equiv 5(\bmod 11) \text {, } \\
A_{2}=\overline{b_{8} b_{9} \cdots b_{68}} \equiv \sum_{i=8}^{68} b_{i} \\
\equiv \sum_{k=0}^{2} \sum_{j=1}^{24} b_{24 k+j}-\sum_{j=1}^{7} b_{j}-\sum_{j=21}^{24} b_{48+j} \\
\equiv \sum_{k=0}^{2}(4 k+2)-0-\sum_{j=21}^{24}\left(70+a_{j}\right) \\
\equiv 18-4 \times 70-\sum_{j=11}^{24} a_{j} \\
\equiv 2-31 \equiv 4(\bmod 11) \text {, } \\
A_{3}=\overline{b_{69} b_{70} \cdots b_{686}} \equiv \sum_{i=69}^{686}(-1)^{i} b_{i} \\
\equiv \sum_{j=21}^{24}(-1)^{j} b_{48+j}+\sum_{k=3}^{27} \sum_{j=1}^{24}(-1)^{j} b_{2 k+j}+ \\
\sum_{j=1}^{14}(-1)^{j} b_{24 \times 28+j} \\
\equiv \sum_{j=21}^{24}(-1)^{j} a_{j}+5 \times 25+\sum_{j=1}^{14}(-1)^{j} a_{j} \\
\equiv 5 \times 26-\sum_{j=15}^{20}(-1)^{j} a_{j} \\
\equiv 5 \times 26-5 \equiv 4(\bmod 11) \text {, } \\
A_{4}=\overline{b_{687} b_{688} \cdots b_{6857}} \equiv \sum_{i=687}^{6857} b_{i} \equiv \sum_{i=687}^{785} b_{i} \\
\equiv \sum_{j=15}^{24} b_{24 \times 28+j}+\sum_{k=29}^{31} \sum_{j=1}^{24} b_{24 k+j}+\sum_{j=1}^{17} b_{24 \times 32+j} \\
=\sum_{j=15}^{24}\left(35 \times 28+a_{j}\right)+\sum_{k=2}^{31}(4 k+2)+\sum_{j=1}^{17}\left(35 \times 32+a_{j}\right) \\
=1+\sum_{j=1}^{24} a_{j}+\sum_{j=15}^{17} a_{j} \\
\equiv 1+2+3 \equiv 6(\bmod 11) \text {. } \\
\end{array}
$$

Therefore, $A \equiv \sum_{i=1}^{4} A_{i} \equiv 8(\bmod 11)$.