Question

The sum of a two digit number and the number obtained by interchanging the digits is 99. If the digits differ by 3. find the number.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{digit}\mathrm{at}\mathrm{units}\mathrm{place}\mathrm{be}x\mathrm{and}\mathrm{the}\mathrm{digit}\text{at}\mathrm{tens}\mathrm{place}\mathrm{be}y. \\ \mathrm{The}\mathrm{number}\mathrm{is}10y+x. \\ \mathrm{The}\mathrm{number}\mathrm{obtained}\mathrm{by}\mathrm{interchanging}\mathrm{the}\mathrm{digits}\mathrm{is}10x+y. \\ \mathrm{Sum}\mathrm{of}\mathrm{these}\mathrm{two}\mathrm{numbers}\mathrm{is}99. \\ \Rightarrow 10y+x+10x+y=99 \\ \Rightarrow 11x+11y=99 \\ \Rightarrow 11\left(x+y\right)=99 \\ \Rightarrow x+y=9...\left(\mathrm{i}\right) \\ \mathrm{The}\mathrm{digits}\mathrm{differ}\mathrm{by}3;\mathrm{so},\mathrm{there}\mathrm{will}\mathrm{be}\mathrm{two}\mathrm{case}. \\ \mathrm{Case}\left(1\right): \\ x>y \\ \mathrm{As}\mathrm{per}\mathrm{case}\left(1\right),x-y=3...\left(\mathrm{ii}\right) \\ \text{By}\mathrm{adding}\left(\mathrm{i}\right)\text{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}: \\ x+y=9 \\ \underline{+x-y=3} \\ 2x=12 \\ \therefore x=6 \\ \mathrm{Substituting}x=6\mathrm{in}x+y=9,\text{we get:} \\ 6+y=9 \\ \therefore y=3 \\ \text{So},\text{t}\mathrm{he}\mathrm{two}\mathrm{digit}\mathrm{number}=10y+x \\ =10\times 3+6 \\ =36 \\ \mathrm{Case}\left(2\right): \\ x<y \\ \mathrm{As}\mathrm{per}\mathrm{case}\left(2\right),y-x=3 \\ -x+y=3...\left(\mathrm{iii}\right) \\ \mathrm{By}\mathrm{adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}: \\ x+y=9 \\ \underline{-x+y=3} \\ 2y=12 \\ \therefore y=6 \\ \mathrm{Substituting}y=6\mathrm{in}x+y=9,\text{we get}: \\ x+6=9 \\ \therefore x=3 \\ \text{So, t}\mathrm{he}\mathrm{two}\mathrm{digit}\mathrm{number}=10y+x \\ =10\times 6+3 \\ =63 \end{gathered}$$

Therefore, the two digit number is 36 or 63