To solve the problem, we need to find the probability that a 9-digit number formed using the digits 1 to 9 without repetition is divisible by 11. We will follow these steps:
### Step 1: Understanding the divisibility rule for 11
For a number to be divisible by 11, the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions must be divisible by 11.
Let:
- \( A \) be the sum of the digits in the odd positions: \( A_1 + A_3 + A_5 + A_7 + A_9 \)
- \( B \) be the sum of the digits in the even positions: \( A_2 + A_4 + A_6 + A_8 \)
We need to ensure that \( |A - B| \) is divisible by 11.
### Step 2: Calculate the total sum of digits
The digits used are from 1 to 9. The total sum of these digits is:
\[
A + B = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
\]
### Step 3: Setting up equations
From the divisibility rule, we have:
1. \( A - B \) can be \( 0, 11, -11, 22, -22, \ldots \)
2. Since \( A + B = 45 \), we can express \( B \) in terms of \( A \):
\[
B = 45 - A
\]
Thus, substituting into the first equation gives:
\[
A - (45 - A) = 2A - 45
\]
We need \( |2A - 45| \) to be divisible by 11.
### Step 4: Finding valid values for \( A \)
Setting \( 2A - 45 = k \cdot 11 \) for some integer \( k \):
- For \( k = 0 \): \( 2A = 45 \) → \( A = 22.5 \) (not valid)
- For \( k = 1 \): \( 2A - 45 = 11 \) → \( 2A = 56 \) → \( A = 28 \)
- For \( k = -1 \): \( 2A - 45 = -11 \) → \( 2A = 34 \) → \( A = 17 \)
Thus, the valid sums for \( A \) are \( 28 \) and \( 17 \).
### Step 5: Finding combinations of digits
1. **For \( A = 28 \)**:
- The digits that sum to 28 can be \( 1, 2, 5, 9 \) (sum = 17) and permutations of these.
- The remaining digits (4 digits) can be arranged in odd positions.
2. **For \( A = 17 \)**:
- The digits that sum to 17 can be \( 1, 2, 6, 8 \) (sum = 17) and permutations of these.
- The remaining digits (4 digits) can be arranged in odd positions.
### Step 6: Counting arrangements
Each valid combination can be arranged in:
- \( 4! \) ways for the digits in odd positions.
- \( 5! \) ways for the remaining digits.
Thus, for each valid sum, the total arrangements are:
\[
\text{Total arrangements} = 11 \times 5! \times 4!
\]
### Step 7: Total arrangements and probability
The total number of arrangements of 9 digits is \( 9! \). Therefore, the probability \( P \) that a randomly formed number is divisible by 11 is:
\[
P = \frac{11 \times 5! \times 4!}{9!}
\]
### Step 8: Simplifying the probability
Calculating \( P \):
\[
P = \frac{11 \times 120 \times 24}{362880} = \frac{11 \times 2880}{362880} = \frac{11}{126}
\]
### Step 9: Finding \( p \) and \( q \)
Here, \( p = 11 \) and \( q = 126 \). We need to find \( \frac{p + q}{100} \):
\[
p + q = 11 + 126 = 137
\]
Thus,
\[
\frac{p + q}{100} = \frac{137}{100} = 1.37
\]
### Step 10: Greatest integer function
Finally, applying the greatest integer function:
\[
\lfloor 1.37 \rfloor = 1
\]
### Final Answer
The answer is:
\[
\boxed{1}
\]
