Counting Four-Digit Numbers with Even Digits

The question asks us to find the total number of four-digit natural numbers where all the digits used are even.

Understanding the Requirements

A natural number is a positive whole number (1, 2, 3, ...). A four-digit number is a natural number between 1000 and 9999, inclusive.

The digits that are considered even are: 0, 2, 4, 6, 8. There are 5 even digits in total.

Applying Constraints to Digits

Let's consider the four positions of a four-digit number:

• Thousands place
• Hundreds place
• Tens place
• Units place

We need to determine the number of choices for each place, keeping in mind the constraints:

1. The number must be a four-digit number, meaning the first digit (thousands place) cannot be 0.
2. All digits must be even (from the set {0, 2, 4, 6, 8}).

Calculating Choices for Each Digit

• Thousands place: This digit must be even, but it cannot be 0. So, the possible choices are {2, 4, 6, 8}. This gives us 4 possible choices.
• Hundreds place: This digit can be any of the even digits. The possible choices are {0, 2, 4, 6, 8}. This gives us 5 possible choices.
• Tens place: Similar to the hundreds place, this digit can be any of the even digits {0, 2, 4, 6, 8}. This gives us 5 possible choices.
• Units place: This digit can also be any of the even digits {0, 2, 4, 6, 8}. This gives us 5 possible choices.

Total Count Calculation

To find the total number of such four-digit natural numbers, we multiply the number of choices for each digit together. This is based on the fundamental principle of counting.

Total count = (Choices for Thousands) \(\times\) (Choices for Hundreds) \(\times\) (Choices for Tens) \(\times\) (Choices for Units)

Total count = \(4 \times 5 \times 5 \times 5\)

Total count = \(4 \times (5^3)\)

Total count = \(4 \times 125\)

Total count = 500

Conclusion

Therefore, there are 500 four-digit natural numbers where all the digits are even.