Question

The sum of the digits of a $2-digit$ number is $9$. The number is $6$ times the unit digit. Find the number.

Answer 1

Let the digit in the tens place be $x$ and the digit in the $y$

The number is $10x+y$

The sum of the digits is $9$

$x+y=9$

$\Rightarrow y=9-x$

The number is $6$ times the units place

$10x+y=6y$

Put the two equations together

$x=9-y$ .....$\left(1\right)$

$10x+y=6y$ .....$\left(2\right)$

Eqn$\left(2\right)-$Eqn$\left(1\right)$ we get

$10(9-y)+y=6y$

$\Rightarrow 90-10y+y=6y$

$\Rightarrow 90-9y=6y$

$\Rightarrow 15y=90$

$\Rightarrow y=\frac{90}{15}=6$

When $y=6$ then $x=9-6=3$

$\therefore$ the number $=10x+y=10\times 3+6=30+6=36$