The sum of two digit number …

Let's denote the tens digit of the two-digit number as \( x \) and the units digit as \( y \). According to the given conditions: 1. The sum of the digits is 9: \[ x + y = 9 \] 2. The number is 6 times the unit digit: \[ 10x + y = 6y \] Now, we have a system of two equations: \[ \begin{cases} x + y = 9 \\ 10x + y = 6y \end{cases} \] From the first equation, we can express \( x \) in terms of \( y \): \[ x = 9 - y \] Substitute this expression for \( x \) into the second equation: \[ 10(9 - y) + y = 6y \] \[ 90 - 10y + y = 6y \] \[ 90 - 9y = 6y \] \[ 90 = 15y \] \[ y = 6 \] Now, substitute the value of \( y \) back into the first equation to find \( x \): \[ x = 9 - 6 \] \[ x = 3 \] So, the tens digit \( x \) is 3 and the units digit \( y \) is 6. Therefore, the number is 36.