Solution:
To determine how many four-digit natural numbers exist such that all of the digits are even, we first need to identify the possible even digits. The even digits are 0,2,4,6, and 8 .
A four-digit natural number ranges from 1000 to 9999 . Therefore, the first digit (thousands place) cannot be 0 . This gives us the even digits 2,4 , 6 , and 8 as options for the first digit.
For each of the remaining three digits (hundreds place, tens place, and units place), any of the even digits (0,2,4,6,8) can be used.
Thus, we have the following choices for each place value in the four-digit number:
Thousands place: 4 choices (2,4,6,8)
Hundreds place: 5 choices (0,2,4,6,8)
Tens place: 5 choices (0,2,4,6,8)
Units place: 5 choices (0,2,4,6,8)
The total number of four-digit natural numbers can then be calculated by multiplying the number of choices for each digit:
4×5×5×5
Calculating this, we get:
4×53=4×125=500
Therefore, there are 500 four-digit natural numbers such that all of the digits are even.
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