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.