If we look at the number $a_6a_5a_4a_3a_2a_1a_0$:
$$a_6a_5a_4a_3a_2a_1a_0 = a_610^6+a_510^5+ \dots +a_010^0 \\ \equiv a_6(-1)^6 + \dots+a_0(-1)^0 \mod 11 = a_6 -a_5+a_4-a_3+a_2-a_1+a_0 \mod 11$$
So if the number will be divisible by 11 we require
$$a_6 -a_5+a_4-a_3+a_2-a_1+a_0 = 11m$$
We take note that $0 \leq a_i \leq 9$ so possible values of $11m$ are limited to:
$$-27 = 4\cdot0-3\cdot9\leq 11m \leq 4\cdot 9-3\cdot0 = 36$$
Which really means that $11m \in \{-22,-11,0,11,22,33\}$
The other requirement of the question was that $a_6 + \dots +a_0 = 59$. From this and the above equation (not sure how to number them and align them nicely in TeX) we add and subtract and get much nicer equations:
$$a_6 +a_4+a_2+a_0 = \frac{59+11m} 2
\\a_5+a_3+a_1 =\frac{ 59-11m}2$$
Of course the LHS is whole, so the right hand side must be as well, which means $m$ needs to be odd. So we reduce our options to:
$$11m \in \{-11,11,33\} \implies \frac{59+11m} 2 \in \{24,35,46\}$$
Of course the sum of four digits can't be $46$ from our above inequality on $a_i$, and similarly
$$\frac{59-11m} 2 \in \{35,24\}$$ But the sum of three digits can't be $35$, so we're left with
$$a_6 +a_4+a_2+a_0 = 35
\\a_5+a_3+a_1 =24$$
It's easy to see that the only options for the four digits is a permutation of $9998$, and the three digits must be a permutation of one of $\{699,789,888\}$.
Order doesn't matter, so basic combinatorics gives $4\cdot(3 + 3! + 1) = 40$ such numbers.
A refinement of your first approach:
In numbers which end with $2$, the sum of the other two digits must be $4$ or $7$:
- $132$
the sum of the other two digits is 4
- $312$
the sum of the other two digits is 4
- $342$
the sum of the other two digits is 7
- $432$
the sum of the other two digits is 7
In numbers which end with $4$, the sum of the other two digits must be $5$ or $8$:
- $234$
the sum of the other two digits is 5
- $324$
the sum of the other two digits is 5
- $354$
the sum of the other two digits is 8
- $534$
the sum of the other two digits is 8
Best Answer
Let $N_7 = \text{what you are looking for}$
However, this consists of one set that doesn't use any zeros - $(NZ)_7$ - and another that does. The number in the set that uses zeros can be defined recursively as $6(NZ)_6$ because the zero can be placed in 6 different places.
Notice that $(NZ)_7 = \frac{\text{pair of two numbers that are divisible by 3}}{9c2} \cdot(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3)$
This is because the total sum of $1,2, ..8, 9$ is $45$ and we need to remove two numbers such that the sum of the others is still a multiple fo $3$ ie the two numbers we remove must be a multiple of $3$
Therefore, $(NZ)_7 = \frac{(NZ)_2}{9P2}(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3)$
Similarly, $(NZ)_6 = \frac{(NZ)_3}{9P3}(9\cdot8\cdot7\cdot6\cdot5\cdot4)$
$(NZ)_2$ can be counted, but also it can be expressed as $9\cdot3 - \frac{1}{9}\cdot 9 \cdot 3$. This is found because if the first is $0 \mod 3$, then there is an overcounting (1/3 of such cases are false $1/3 *1/3 = 1/9$)
Similarly, $(NZ)_2=9\cdot 8 \cdot 3 - \frac{2}{8} \cdot \frac{2}{3} \cdot 9 \cdot 8\cdot 3$. Here, the second part is found by considering the probability that the second number chosen is the same modulus as the first ( $2/8$), which results in only $1/3$ of the possible last digits (so we subtract out $2/3$)
Therefore,
$$N_7 = \frac{9 \cdot 3 - 3}{72}(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3) + 6 \cdot \frac{9\cdot8\cdot3-4\cdot9}{504}(9\cdot8\cdot7\cdot6\cdot5\cdot4) = 190080$$
This is definitely not my most eloquent answer, so please ask me questions if something is confusing, and I'll try to explain my thought process.