To solve this problem, we need to count the number of natural numbers less than 7000 that can be formed using the digits 0, 1, 3, 7, and 9, with repetition allowed. Let's break this down by considering numbers with different digit counts:
**1-digit numbers:**
The possible 1-digit numbers are 1, 3, 7, and 9. Thus, there are 4 possible numbers.
**2-digit numbers:**
The first digit can be 1, 3, 7, or 9 (as it cannot be 0), giving us 4 options. Each subsequent digit has 5 options (including 0). Hence, there are 4 \times 5 = 20 possible 2-digit numbers.
**3-digit numbers:**
Similarly, the first digit can be 1, 3, 7, or 9, giving 4 choices. Each of the remaining two places can be filled with 5 options each. Therefore, there are 4 \times 5 \times 5 = 100 possible 3-digit numbers.
**4-digit numbers less than 7000:**
The thousands digit can be 0, 1, 3, or 6 (it cannot be 7 or more). Thus, we have 3 options for the first digit. Each of the remaining digits (hundreds, tens, units) can again be any of the 5 digits. Therefore, there are 4 \times 5 \times 5 \times 5 = 125 possible 4-digit numbers.
Summing these possibilities gives us the total number of potential numbers:
4 + 20 + 100 + 250 = 374
Therefore, the number of natural numbers less than 7000 that can be formed by using the digits 0, 1, 3, 7, and 9 is 374. The correct answer is 374.
