Understanding the Problem
To find the sum of all 4-digit numbers formed by the digits 2, 4, 6, and 8 without repetition, we will follow a systematic approach.
Counting the Possible Numbers
- Each digit can be used once, leading to permutations of the 4 digits.
- The total number of permutations is 4! (factorial of 4), which equals 24.
Calculating Contribution of Each Digit
- Each digit (2, 4, 6, 8) will appear in each place (thousands, hundreds, tens, units) equally across all permutations.
- Since there are 24 total numbers, each digit will appear in each position (thousands, hundreds, tens, units) 6 times (24 total / 4 digits).
Calculating the Total Contribution
- The contribution of each digit in each place is calculated as follows:
- Thousands Place: Each digit contributes to 1000 times its value.
- Hundreds Place: Each digit contributes to 100 times its value.
- Tens Place: Each digit contributes to 10 times its value.
- Units Place: Each digit contributes to its own value.
Final Calculation
- Total contribution of each digit:
- 2 * (1000 + 100 + 10 + 1) * 6
- 4 * (1000 + 100 + 10 + 1) * 6
- 6 * (1000 + 100 + 10 + 1) * 6
- 8 * (1000 + 100 + 10 + 1) * 6
- Simplifying:
- Total Value per Digit = (1111) * 6 = 6666
- Multiply by the sum of the digits (2 + 4 + 6 + 8 = 20).
Total Sum of All 4-Digit Numbers
- Total Sum = 6666 * 20 = 133320.
Thus, the sum of all 4-digit numbers formed by the digits 2, 4, 6, and 8 without repetition is 133320.