Find the sum of all possible 4-digit numbers that can be formed using digits 1, 2, 3, and 4 (repetition allowed).

To solve this problem, we need to find the sum of all possible 4-digit numbers that can be formed using the digits 1, 2, 3, and 4, where repetition of digits is allowed.

### Step-by-Step Solution:

1. **Number of possible 4-digit numbers**:  
   Since repetition of digits is allowed, for each of the four digits (thousands, hundreds, tens, and units), we have 4 possible choices (1, 2, 3, or 4). Therefore, the total number of possible 4-digit numbers is:
   \[
   4 \times 4 \times 4 \times 4 = 256
   \]

2. **Contribution of each digit to the sum**:  
   Let's examine how each digit (1, 2, 3, or 4) contributes to the total sum. Consider the thousands place first. Each digit will appear in the thousands place in exactly one-fourth of all numbers, because every digit has an equal probability of being in any position.

   Since there are 256 total numbers, each digit will appear in the thousands place \( \frac{256}{4} = 64 \) times. The same applies for the hundreds, tens, and units places.

3. **Sum of digits for each place**:  
   The sum of the digits 1, 2, 3, and 4 is:
   \[
   1 + 2 + 3 + 4 = 10
   \]
   Therefore, the contribution of each place to the total sum is calculated as follows:
   
   - **Thousands place**:  
     Each of the digits 1, 2, 3, and 4 appears 64 times in the thousands place, and their sum is multiplied by 1000 (since it's the thousands place):
     \[
     64 \times 10 \times 1000 = 640,000
     \]
   
   - **Hundreds place**:  
     Similarly, the contribution of the hundreds place is:
     \[
     64 \times 10 \times 100 = 64,000
     \]
   
   - **Tens place**:  
     The contribution of the tens place is:
     \[
     64 \times 10 \times 10 = 6,400
     \]
   
   - **Units place**:  
     The contribution of the units place is:
     \[
     64 \times 10 \times 1 = 640
     \]

4. **Total sum**:  
   Now, we sum up the contributions from all four places:
   \[
   640,000 + 64,000 + 6,400 + 640 = 711,040
   \]

### Final Answer:
The sum of all possible 4-digit numbers that can be formed using the digits 1, 2, 3, and 4 is \( \boxed{711,040} \).

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Would you like more details or have any further questions?

Here are 5 related questions to expand on this concept:

1. How would the sum change if repetition was not allowed?
2. How does the total number of 4-digit numbers change if we introduce more digits?
3. What would be the sum of all 3-digit numbers formed using the digits 1, 2, and 3?
4. How would the sum change if we used the digits 0, 1, 2, and 3 instead?
5. What is the average of all the 4-digit numbers formed?

**Tip**: In problems involving sums of digits in different places, it helps to calculate the contribution of each place separately for clarity.

Find the sum of all 4-digit numbers formed using digits 1, 2, 3, and 4, with repetition allowed. Learn how to calculate the contribution of each place value and apply principles of combinatorics and arithmetic. Suitable for students in Grades 9-12.

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