Find the number of $7$-digit numbers, where the sum of the digits is divisible by $11.$

Answer 1

To find the number of \(7\)-digit numbers where the sum of the digits is divisible by \(11\), we need to approach the problem methodically. A \(7\)-digit number can be represented as \( \overline{a_1a_2a_3a_4a_5a_6a_7} \), where \( a_1, a_2, \ldots, a_7 \) are the digits and \( a_1 \neq 0 \) (since it's a 7-digit number).

First, the requirements for a \(7\)-digit number indicate:
\[ 1 \leq a_1 \leq 9 \quad \text{and} \quad 0 \leq a_2, a_3, a_4, a_5, a_6, a_7 \leq 9 \]

We aim to count cases where the sum \( S = a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 \) is divisible by \(11\).

### Step-by-Step Solution:

1. **Count the Total Number of \(7\)-Digit Numbers:**
\[9\] choices for \(a_1\) (since \(a_1 \neq 0\)) and \(10\) choices for each of \(a_2, a_3, a_4, a_5, a_6, a_7\) give:
\[
9 \times 10^6
\]
total \(7\)-digit numbers.

2. **Distribution of Sums:**
\( a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 \) ranges from \(1\) (minimum) to \(63\) (maximum). Now, the sum \( S \) being divisible by \(11\) means:
\[ S \equiv 0 \pmod{11} \]

3. **Check for Modular Conformity:**
Possible sums \( S \) such that \( S \equiv 0 \pmod{11} \) within the range \( [1, 63] \) are:
\[
11, 22, 33, 44, 55
\]

4. **Probability of Satisfying Divisibility:**
Given uniform distributions of digits, each sum condition \( \sum a_i = k \mod 11\) should be equally likely, recognizing \(0 \pmod{11} \) as equally probable for sums \(0, 1, \ldots, 10\). Thus, exactly \( \frac{1}{11} \) of numbers will satisfy \( S \equiv 0 \pmod{11} \).

5. **Calculate Required Count:**
Since we need the fraction \(1/11\) of the total \(7\)-digit numbers, calculate:
\[
\frac{1}{11} \times 9 \times 10^6 = \frac{9,000,000}{11} \approx 818,182 \quad (exactly)
\]

### Conclusion:

The number of \(7\)-digit numbers where the sum of the digits is divisible by \(11\) is:
\[
\boxed{818182}
\]