To solve this problem, we need to find the sum of all four-digit numbers that can be formed using the digits 2, 4, 6, and 8, where each digit is used exactly once in each arrangement.

Step 1: Number of possible arrangements

The digits given are 2, 4, 6, and 8. Since all four digits are distinct and must be used exactly once, the number of different four-digit numbers is the number of permutations of 4 digits, which is:

$4! = 4 \times 3 \times 2 \times 1 = 24$

So, there are 24 possible four-digit numbers.

Step 2: Contribution of each digit

Each of the four digits (2, 4, 6, and 8) will appear in each place (thousands, hundreds, tens, and units) an equal number of times. Since there are 24 numbers and 4 digits, each digit will appear in each place:

$\frac{24}{4} = 6 \text{ times in each place}.$

Step 3: Contribution to the total sum

Now, let’s consider the total contribution of each digit. Each digit appears 6 times in the thousands, hundreds, tens, and units place. The value of the contribution of a digit to the total sum depends on its position. Specifically:

• A digit in the thousands place contributes its value multiplied by 1000.
• A digit in the hundreds place contributes its value multiplied by 100.
• A digit in the tens place contributes its value multiplied by 10.
• A digit in the units place contributes its value directly (multiplied by 1).

Thus, the total contribution of a digit $d$ to the sum is:

$d \times (1000 + 100 + 10 + 1) = d \times 1111.$

Each digit (2, 4, 6, 8) will contribute $d \times 1111$ to the total sum. Since each digit appears 6 times, the total contribution of each digit is:

$6 \times d \times 1111.$

Step 4: Total sum of all numbers

Now, we compute the total contribution for each digit:

• For $d = 2$: $6 \times 2 \times 1111 = 13332$
• For $d = 4$: $6 \times 4 \times 1111 = 26664$
• For $d = 6$: $6 \times 6 \times 1111 = 39996$
• For $d = 8$: $6 \times 8 \times 1111 = 53328$

Finally, summing these values gives the total sum of all numbers:

$13332 + 26664 + 39996 + 53328 = 133320.$

Final Answer:

The sum of all four-digit numbers that can be formed using the digits 2, 4, 6, and 8 is:

$\boxed{133320}.$

Would you like further details or clarifications on any of the steps?

Here are 5 related questions to extend this problem:

1. How many distinct three-digit numbers can be formed using the digits 1, 3, and 5?
2. What is the sum of all three-digit numbers formed using the digits 1, 3, and 5?
3. How does the formula for the sum change if we have more digits or if repetition is allowed?
4. What is the sum of all even four-digit numbers that can be formed with the digits 2, 4, 6, and 8?
5. How many five-digit