Find the sum of all those numbers which can be formed with the digits 0,1,2,3 taken together.

Answer

To find the sum of all numbers that can be formed using the digits 0, 1, 2, and 3, we will follow these steps:

### Step 1: Determine the valid numbers
Since we are using the digits 0, 1, 2, and 3, we need to consider the fact that a number cannot start with 0. Therefore, the valid starting digits for our numbers can only be 1, 2, or 3.

### Step 2: Count the total arrangements
We can form numbers of different lengths using these digits. The possible lengths are 1-digit, 2-digit, 3-digit, and 4-digit numbers.

- **1-digit numbers**: The valid numbers are 1, 2, and 3. Total = 3.

- **2-digit numbers**: The first digit can be 1, 2, or 3 (3 choices), and the second digit can be any of the 4 digits (0, 1, 2, 3). Total = 3 * 4 = 12.

- **3-digit numbers**: The first digit can be 1, 2, or 3 (3 choices), the second digit can be any of the 4 digits, and the third digit can also be any of the 4 digits. Total = 3 * 4 * 4 = 48.

- **4-digit numbers**: The first digit can be 1, 2, or 3 (3 choices), and the remaining three digits can be any of the 4 digits. Total = 3 * 4 * 4 * 4 = 192.

### Step 3: Calculate the sum of all the numbers
Now we will calculate the contribution of each digit in each position (units, tens, hundreds, thousands) for the 2-digit, 3-digit, and 4-digit numbers.

#### Contribution of each digit:
1. **1-digit numbers**:
- Sum = 1 + 2 + 3 = 6.

2. **2-digit numbers**:
- Each digit appears in the tens place and units place.
- Each of the digits 1, 2, and 3 appears in the tens place for 4 times (as the second digit can be any of the 4 digits).
- Contribution from tens place = (1 + 2 + 3) * 10 * 4 = 6 * 10 * 4 = 240.
- Each digit appears in the units place for 3 times (as the first digit can be 1, 2, or 3).
- Contribution from units place = (0 + 1 + 2 + 3) * 3 = 6 * 3 = 18.
- Total for 2-digit numbers = 240 + 18 = 258.

3. **3-digit numbers**:
- Each digit appears in the hundreds, tens, and units places.
- Contribution from hundreds place = (1 + 2 + 3) * 100 * 16 = 6 * 100 * 16 = 9600.
- Contribution from tens place = (0 + 1 + 2 + 3) * 4 * 4 * 3 = 6 * 4 * 3 = 72.
- Contribution from units place = (0 + 1 + 2 + 3) * 4 * 4 = 6 * 16 = 96.
- Total for 3-digit numbers = 9600 + 72 + 96 = 9786.

4. **4-digit numbers**:
- Each digit appears in the thousands, hundreds, tens, and units places.
- Contribution from thousands place = (1 + 2 + 3) * 1000 * 64 = 6 * 1000 * 64 = 384000.
- Contribution from hundreds place = (0 + 1 + 2 + 3) * 16 * 4 * 3 = 6 * 16 * 3 = 288.
- Contribution from tens place = (0 + 1 + 2 + 3) * 16 * 4 = 6 * 16 = 96.
- Contribution from units place = (0 + 1 + 2 + 3) * 16 = 6 * 16 = 96.
- Total for 4-digit numbers = 384000 + 288 + 96 + 96 = 384480.

### Step 4: Final sum
Now we will sum all contributions:
- Total = 6 (1-digit) + 258 (2-digit) + 9786 (3-digit) + 384480 (4-digit) = 394530.

Thus, the sum of all the numbers that can be formed with the digits 0, 1, 2, and 3 is **394530**.