Question

The sum of the digits of a two-digit number is 9. Nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Hint:

When solving problems involving two-digit numbers and conditions on the digits, use variables to represent the digits, form equations based on the given conditions, and solve the system of equations.

Answer 1

Let the two-digit number be \( 10x + y \), where \( x \) is the tens digit and \( y \) is the ones digit. The sum of the digits is 9, so we have: \[ x + y = 9 \quad \text{(Equation 1)}.
\] Next, we are told that nine times the number is twice the number obtained by reversing the digits. The number obtained by reversing the digits is \( 10y + x \), so we can write: \[ 9(10x + y) = 2(10y + x).
\] Simplifying the equation: \[ 90x + 9y = 20y + 2x \quad \Rightarrow \quad 90x - 2x = 20y - 9y \quad \Rightarrow \quad 88x = 11y.
\] Now divide both sides by 11: \[ 8x = y.
\]
Step 1: Substitute \( y = 8x \) into Equation 1: \[ x + 8x = 9 \quad \Rightarrow \quad 9x = 9 \quad \Rightarrow \quad x = 1.
\]
Step 2: Now substitute \( x = 1 \) into \( y = 8x \): \[ y = 8 \times 1 = 8.
\] Thus, the number is \( 10x + y = 10(1) + 8 = 18.
\)
Conclusion: The number is \( 18 \).