To solve the problem of forming an eight-digit number divisible by 9 using the digits from 0 to 9 without repeating any digits, we can follow these steps:
### Step 1: Understand the divisibility rule for 9
A number is divisible by 9 if the sum of its digits is divisible by 9. The sum of the digits from 0 to 9 is \(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45\), which is divisible by 9.
### Step 2: Determine the requirement for an 8-digit number
To form an 8-digit number, we need to remove 2 digits from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. After removing 2 digits, the sum of the remaining digits must still be divisible by 9.
### Step 3: Identify pairs of digits that can be removed
We need to find pairs of digits whose sum is 9, since removing them will keep the remaining sum (which is 45) divisible by 9. The possible pairs are:
- (0, 9)
- (1, 8)
- (2, 7)
- (3, 6)
- (4, 5)
### Step 4: Calculate the number of valid 8-digit numbers
For each pair removed, we can form an 8-digit number. The total number of ways to arrange the remaining 8 digits is \(8!\). However, we must ensure that the first digit is not 0 (as it cannot be the leading digit in an 8-digit number).
#### Case 1: Remove the pair (0, 9)
- Remaining digits: {1, 2, 3, 4, 5, 6, 7, 8}
- Total arrangements: \(8!\)
#### Case 2: Remove the pair (1, 8)
- Remaining digits: {0, 2, 3, 4, 5, 6, 7}
- Total arrangements: \(7!\) (since 0 cannot be the first digit)
#### Case 3: Remove the pair (2, 7)
- Remaining digits: {0, 1, 3, 4, 5, 6, 8}
- Total arrangements: \(7!\) (since 0 cannot be the first digit)
#### Case 4: Remove the pair (3, 6)
- Remaining digits: {0, 1, 2, 4, 5, 7, 8}
- Total arrangements: \(7!\) (since 0 cannot be the first digit)
#### Case 5: Remove the pair (4, 5)
- Remaining digits: {0, 1, 2, 3, 6, 7, 8}
- Total arrangements: \(7!\) (since 0 cannot be the first digit)
### Step 5: Combine the cases
The total number of valid arrangements is:
- For (0, 9): \(8!\)
- For (1, 8): \(7!\)
- For (2, 7): \(7!\)
- For (3, 6): \(7!\)
- For (4, 5): \(7!\)
Thus, the total number of valid 8-digit numbers is:
\[
8! + 4 \times 7!
\]
### Step 6: Simplify the expression
Calculating \(8!\) and \(7!\):
- \(8! = 40320\)
- \(7! = 5040\)
Now substituting these values:
\[
Total = 40320 + 4 \times 5040 = 40320 + 20160 = 60480
\]
### Step 7: Final answer
The total number of ways to form an 8-digit number divisible by 9 using the digits from 0 to 9 without repeating any digits is \(36 \times 7!\).
