A $\displaystyle 2$-digit number is seven times the sum of its digits. The number formed by reversing the digits is $\displaystyle 18$ less than the given number. Find the given number.

Let unit's digit be x and ten's digit be y. ∴ Number \(\displaystyle \boldsymbol{=} \mathbf{1 0 y} \boldsymbol{+} \mathbf{x}\) \(\displaystyle \mathbf{1 0 y}+\mathbf{x}=\mathbf{7}(\mathbf{x}+\mathbf{y}) \Rightarrow 3 \mathbf{y}-\mathbf{6 x}=\mathbf{0}\) \(\displaystyle \mathrm{y}=2 \mathrm{x} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(\mathrm{i})\) \(\displaystyle 10 y+x-(10 x+y)=18\) \(\displaystyle 9 y-9 x=18\) or \(\displaystyle y-x=2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \ldots .\). (ii) On solving (i) and (ii), \(\displaystyle \mathrm{x}=2, \mathrm{y}=4\) ∴ required number is $\displaystyle 42$