Let the number be \( 10a + b \), where \( a \) is the tens digit and \( b \) is the ones digit.
Step 1: According to the problem, the sum of the digits is 9, so: \[ a + b = 9 \]
Step 2: When 27 is added to the number, the digits are reversed. So, we have the equation: \[ 10a + b + 27 = 10b + a \]
Step 3: Simplify the equation: \[ 10a + b + 27 = 10b + a \implies 9a - 9b = -27 \] \[ a - b = -3 \]
Step 4: Solve the system of equations: 1. \( a + b = 9 \) 2. \( a - b = -3 \) Add both equations: \[ 2a = 6 \implies a = 3 \] Substitute \( a = 3 \) into \( a + b = 9 \): \[ 3 + b = 9 \implies b = 6 \] Thus, the number is \( 10a + b = 10(3) + 6 = 36 \).