Question

A two-digit number is obtained by either multiplying the sum of the digits by $8$ and adding $1$ or by multiplying the difference of the digits by $13$ and adding $2$. Find the number.

Answer 1

Let the digit at units place be x and the digit at ten's place be y. Then,
Number $=10y+x$

According to the given conditions, we have

$10y+x=8(x+y)+1\Rightarrow 7x-2y+1=0$

and, $10y+x=13(y-x)+2\Rightarrow 14x-3y-2=0$

By using cross-multiplication, we have

$\frac{x}{-2\times -2-(-3)\times 1}=\frac{-y}{7\times -2-14\times 1}=\frac{1}{7\times -3-14\times -2}$

$\Rightarrow \frac{x}{4+3}=\frac{-y}{-14-14}=\frac{1}{-21+28}$

$\Rightarrow \frac{x}{7}=\frac{y}{28}=\frac{1}{7}$

$\Rightarrow x=\frac{7}{7}=1$ and $y=\frac{28}{7}=4$

Hence, the number $=10y+x=10\times 4+1=41$.