Question

The sum of digits in a three digit number is nine. The tens digit is half the sum of other two and the hundreds digits is half the units digit. Find the number

Answer 1

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:

1. The sum of the digits is 9:
\[
a + b + c = 9
\]

2. The tens digit is half the sum of the other two digits:
\[
b = \frac{1}{2}(a + c)
\]

3. 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 \( b \):
1. \( b = 9 - 3a \)
2. \( b = \frac{3}{2}a \)

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 \( \boxed{234} \).