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

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 $\displaystyle 42$