To find the number of three-digit numbers where repetition of digits is allowed and the sum of the digits is even, we can break down the problem into manageable steps.
### Step 1: Understand the conditions
We need to create three-digit numbers (let's denote them as \(abc\)) where:
- \(a\), \(b\), and \(c\) are the digits of the number.
- The sum \(a + b + c\) must be even.
- The first digit \(a\) cannot be 0 (since it is a three-digit number).
### Step 2: Identify even and odd digits
The digits can be classified as:
- Even digits: 0, 2, 4, 6, 8 (5 options)
- Odd digits: 1, 3, 5, 7, 9 (5 options)
### Step 3: Determine cases for the sum to be even
To ensure that the sum \(a + b + c\) is even, we can have the following combinations of digits:
1. All three digits are even.
2. Two digits are odd and one digit is even.
### Case 1: All three digits are even
- The first digit \(a\) can be any of the even digits except 0 (2, 4, 6, 8), giving us 4 choices.
- The second digit \(b\) can be any even digit (0, 2, 4, 6, 8), giving us 5 choices.
- The third digit \(c\) can also be any even digit (0, 2, 4, 6, 8), giving us 5 choices.
Calculating the total for this case:
\[
\text{Total for Case 1} = 4 \times 5 \times 5 = 100
\]
### Case 2: Two digits are odd and one digit is even
We need to consider three sub-cases based on the position of the even digit.
#### Sub-case 2.1: Even digit in the first position
- The first digit \(a\) can be any even digit except 0 (2, 4, 6, 8), giving us 4 choices.
- The second digit \(b\) can be any odd digit (1, 3, 5, 7, 9), giving us 5 choices.
- The third digit \(c\) can also be any odd digit (1, 3, 5, 7, 9), giving us 5 choices.
Calculating the total for this sub-case:
\[
\text{Total for Sub-case 2.1} = 4 \times 5 \times 5 = 100
\]
#### Sub-case 2.2: Even digit in the second position
- The first digit \(a\) can be any odd digit (1, 3, 5, 7, 9), giving us 5 choices.
- The second digit \(b\) can be any even digit (0, 2, 4, 6, 8), giving us 5 choices.
- The third digit \(c\) can also be any odd digit (1, 3, 5, 7, 9), giving us 5 choices.
Calculating the total for this sub-case:
\[
\text{Total for Sub-case 2.2} = 5 \times 5 \times 5 = 125
\]
#### Sub-case 2.3: Even digit in the third position
- The first digit \(a\) can be any odd digit (1, 3, 5, 7, 9), giving us 5 choices.
- The second digit \(b\) can be any odd digit (1, 3, 5, 7, 9), giving us 5 choices.
- The third digit \(c\) can be any even digit (0, 2, 4, 6, 8), giving us 5 choices.
Calculating the total for this sub-case:
\[
\text{Total for Sub-case 2.3} = 5 \times 5 \times 5 = 125
\]
### Step 4: Combine all cases
Now, we can sum the totals from all cases:
\[
\text{Total} = \text{Total for Case 1} + \text{Total for Sub-case 2.1} + \text{Total for Sub-case 2.2} + \text{Total for Sub-case 2.3}
\]
\[
\text{Total} = 100 + 100 + 125 + 125 = 450
\]
### Final Answer
The total number of three-digit numbers where repetition is allowed and the sum of the digits is even is **450**.
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