To solve the problem, we need to find all three-digit numbers (ABC) such that the number itself is equal to the sum of the factorials of its digits. Let's denote the three-digit number as \( N = 100A + 10B + C \), where \( A \), \( B \), and \( C \) are the digits of the number.
### Step 1: Understand the range of digits
Since \( N \) is a three-digit number, \( A \) (the hundreds place) can take values from 1 to 9, while \( B \) and \( C \) (the tens and units places) can take values from 0 to 9.
### Step 2: Factorial values of digits
We need to consider the factorial values of digits from 0 to 9:
- \( 0! = 1 \)
- \( 1! = 1 \)
- \( 2! = 2 \)
- \( 3! = 6 \)
- \( 4! = 24 \)
- \( 5! = 120 \)
- \( 6! = 720 \)
- \( 7! = 5040 \) (not applicable for three-digit numbers)
- \( 8! = 40320 \) (not applicable)
- \( 9! = 362880 \) (not applicable)
### Step 3: Set up the equation
We need to find \( A \), \( B \), and \( C \) such that:
\[
100A + 10B + C = A! + B! + C!
\]
### Step 4: Check possible values
We will check for all combinations of \( A \), \( B \), and \( C \) (where \( A \) ranges from 1 to 9, and \( B \) and \( C \) range from 0 to 9) to see if the equation holds.
1. For \( A = 1 \):
- \( 100 + 10B + C = 1 + B! + C! \)
- Check for \( B \) and \( C \) from 0 to 9.
2. For \( A = 2 \):
- \( 200 + 10B + C = 2 + B! + C! \)
- Check for \( B \) and \( C \).
3. Continue this process for \( A = 3, 4, 5, 6, 7, 8, 9 \).
### Step 5: Identify valid numbers
After checking all combinations, we find:
- \( 145 = 1! + 4! + 5! = 1 + 24 + 120 = 145 \)
- \( 405 = 4! + 0! + 5! = 24 + 1 + 120 = 145 \)
### Step 6: Calculate the sum of valid numbers
The valid three-digit numbers that satisfy the condition are \( 145 \) and \( 405 \). Therefore, the sum \( \lambda \) of all such numbers is:
\[
\lambda = 145 + 405 = 550
\]
### Step 7: Find the sum of the digits of \( \lambda \)
Now, we need to find the sum of the digits of \( \lambda = 550 \):
\[
5 + 5 + 0 = 10
\]
### Final Answer
The sum of the digits of \( \lambda \) is \( 10 \).
