Counting Natural Numbers Less Than 7,000

We need to count the number of natural numbers less than 7,000 that can be formed using the digits \( 0, 1, 3, 7, 9 \) with repetition allowed.

Step 1: Identifying the Digit Constraints

The given digits are: \( 0, 1, 3, 7, 9 \) (five choices per position).

The numbers must be less than 7,000, meaning:

• The first digit can be only 1 or 3 (since 7,000 or more would require 7 or 9).
• The remaining digits can be any of the five available digits.

Step 2: Counting Valid Numbers Based on Number of Digits

1-digit numbers:

Valid choices: \( \{1, 3, 7, 9\} \) (not 0) → Total: \( 4 \).

2-digit numbers:

First digit: \( \{1, 3, 7, 9\} \) (4 choices)
Second digit: \( \{0, 1, 3, 7, 9\} \) (5 choices)
Total: \( 4 \times 5 = 20 \).

3-digit numbers:

First digit: \( 4 \) choices
Second digit: \( 5 \) choices
Third digit: \( 5 \) choices
Total: \( 4 \times 5 \times 5 = 100 \).

4-digit numbers:

First digit: \( \{1, 3\} \) (2 choices, since number must be < 7,000)
Second digit: \( 5 \) choices
Third digit: \( 5 \) choices
Fourth digit: \( 5 \) choices
Total: \( 2 \times 5 \times 5 \times 5 = 250 \).

Step 3: Adding Up the Cases

\[ 4 + 20 + 100 + 250 = 374 \]

Thus, the total number of natural numbers less than 7,000 that can be formed is 374.