Question

The sum of two digit number and number obtained by reversing the order of its digits is 99 if digit differ by 3 find the number

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{digit}\mathrm{at}\mathrm{unit}\text{'}\mathrm{s}\mathrm{place}=x \\ \mathrm{Let}\mathrm{the}\mathrm{digit}\mathrm{at}\mathrm{ten}\text{'}\mathrm{s}\mathrm{place}=y \\ \mathrm{Now}\mathrm{original}\mathrm{number}=10y+x \\ \mathrm{According}\mathrm{to}\mathrm{first}\mathrm{condition}, \\ x-y=3...............\left(1\right) \\ y-x=3...............\left(2\right) \\ \mathrm{When}\mathrm{the}\mathrm{digits}\mathrm{are}\mathrm{interchanged},\mathrm{then}\mathrm{new}\mathrm{number}=10x+y \\ \mathrm{According}\mathrm{to}\mathrm{second}\mathrm{condition}, \\ 10y+x+10x+y=99 \\ \Rightarrow 11x+11y=99 \\ \Rightarrow x+y=9.................\left(3\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}, \\ 2x=12 \\ \Rightarrow x=6 \\ \mathrm{Now},\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}, \\ 6-y=3 \\ -y=-3 \\ y=3 \\ \mathrm{So},\mathrm{original}\mathrm{number}=10\times 3+6=36 \\ \mathrm{Now},\mathrm{adding}\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}, \\ 2y=12 \\ y=6 \\ \mathrm{Now},\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get}, \\ 6-x=3 \\ -x=3-6 \\ -x=-3 \\ x=3 \\ \mathrm{So},\mathrm{original}\mathrm{number}=10\times 6+3=6 \\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{number}\mathrm{can}\mathrm{be}\mathrm{either}36\mathrm{or}63. \end{gathered}$$