How many natural numbers not more than 4300 can be formed with the digits 0, 1, 2, 3, 4 (if repetitions are allowed)?

Options

A. 574
B. 570
C. 575
D. 569

Answer

To solve the problem of how many natural numbers not more than 4300 can be formed with the digits 0, 1, 2, 3, and 4 (with repetitions allowed), we will break it down into cases based on the number of digits in the numbers.

### Step-by-Step Solution:

**Step 1: Count 1-digit numbers.**
- The possible digits are 1, 2, 3, and 4 (0 cannot be used as a natural number).
- Thus, there are **4 one-digit numbers**: {1, 2, 3, 4}.

**Step 2: Count 2-digit numbers.**
- The first digit (tens place) can be 1, 2, 3, or 4 (4 options).
- The second digit (units place) can be 0, 1, 2, 3, or 4 (5 options).
- Therefore, the total number of 2-digit numbers is:
\[ 4 \times 5 = 20 \]

**Step 3: Count 3-digit numbers.**
- The first digit (hundreds place) can be 1, 2, 3, or 4 (4 options).
- The second digit (tens place) can be 0, 1, 2, 3, or 4 (5 options).
- The third digit (units place) can also be 0, 1, 2, 3, or 4 (5 options).
- Therefore, the total number of 3-digit numbers is:
\[ 4 \times 5 \times 5 = 100 \]

**Step 4: Count 4-digit numbers.**
- We need to consider the restriction that the number must be less than or equal to 4300. We will break this down based on the first digit.

1. **If the first digit is 1, 2, or 3:**
- The first digit can be 1, 2, or 3 (3 options).
- The second digit can be 0, 1, 2, 3, or 4 (5 options).
- The third digit can also be 0, 1, 2, 3, or 4 (5 options).
- The fourth digit can also be 0, 1, 2, 3, or 4 (5 options).
- Therefore, the total for these cases is:
\[ 3 \times 5 \times 5 \times 5 = 375 \]

2. **If the first digit is 4:**
- The number must be less than or equal to 4300, so the second digit can only be 0, 1, 2, or 3 (4 options).
- If the second digit is 0, 1, or 2, the third and fourth digits can be anything (5 options each).
- If the second digit is 3, the third digit can only be 0 (1 option), and the fourth digit can be anything (5 options).
- Therefore, the total for this case is:
- For second digit 0, 1, or 2:
\[ 3 \times 5 \times 5 = 75 \]
- For second digit 3:
\[ 1 \times 5 = 5 \]
- Total for first digit 4:
\[ 75 + 5 = 80 \]

**Total for 4-digit numbers:**
\[ 375 + 80 = 455 \]

**Step 5: Combine all counts.**
- Total numbers = 1-digit + 2-digit + 3-digit + 4-digit
\[ 4 + 20 + 100 + 455 = 579 \]

### Final Answer:
The total number of natural numbers not more than 4300 that can be formed with the digits 0, 1, 2, 3, and 4 is **579**.