Let the three-digit number be represented as \( abc \), where \( a \) is the hundreds digit, \( b \) is the tens digit, and \( c \) is the units digit.
From the problem, we have the following conditions:
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The sum of the digits is 9: \[ a + b + c = 9 \]
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The tens digit is half the sum of the other two digits: \[ b = \frac{1}{2}(a + c) \]
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The hundreds digit is half the units digit: \[ a = \frac{1}{2}c \]
Using the third equation \( a = \frac{1}{2}c \), we can express \( c \) in terms of \( a \): \[ c = 2a \]
Now substitute \( c = 2a \) into the first equation: \[ a + b + 2a = 9 \] \[ 3a + b = 9 \] \[ b = 9 - 3a \]
Next, we substitute \( c = 2a \) into the second equation \( b = \frac{1}{2}(a + c) \): \[ b = \frac{1}{2}(a + 2a) = \frac{1}{2}(3a) = \frac{3}{2}a \]
Now we have two expressions for \( b \):
- \( b = 9 - 3a \)
- \( b = \frac{3}{2}a \)
Setting these two expressions equal to each other: \[ 9 - 3a = \frac{3}{2}a \]
To eliminate the fraction, multiply the entire equation by 2: \[ 18 - 6a = 3a \] Combine like terms: \[ 18 = 9a \] \[ a = 2 \]
Now that we have \( a \), we can find \( c \) and \( b \): \[ c = 2a = 2 \times 2 = 4 \] And substitute \( a \) into the expression for \( b \): \[ b = 9 - 3a = 9 - 3 \times 2 = 9 - 6 = 3 \]
Thus, the digits of the number are: \[ a = 2, \quad b = 3, \quad c = 4 \]
The three-digit number is: \[ 234 \]
Finally, let's verify:
- The sum of the digits \( 2 + 3 + 4 = 9 \) (correct).
- The tens digit \( 3 \) is half the sum of the other two: \( \frac{1}{2}(2 + 4) = \frac{1}{2} \times 6 = 3 \) (correct).
- The hundreds digit \( 2 \) is half the units digit \( 4 \) (correct).
Therefore, the three-digit number is \( \boxed{234} \).
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