Question

The sum of the digits of a two-digit number is $9$. When we interchange the digits, it is found that the resulting new number is greater than the original number by $27$. What is the two-digit number?

Answer 1

Finding a two-digit number:

We have to find a two-digit number whose sum of digits is $9$ and when the digits are reversed the new number is greater than the original number by $27$.

Let the digit at ten's place be $\mathrm{x}$.

Then the digit at one's place $=9-\mathrm{x}$ (since the sum of digits is $9$)

Thus the number $=10\mathrm{x}+\left(9-\mathrm{x}\right)$

The number is formed by reversing the digits $=10\left(9-\mathrm{x}\right)+\mathrm{x}$

According to the question,

$$\begin{gathered} 10\left(9-\mathrm{x}\right)+\mathrm{x}=10\mathrm{x}+\left(9-\mathrm{x}\right)+27 \\ \Rightarrow 90-10\mathrm{x}+\mathrm{x}=10\mathrm{x}+9-\mathrm{x}+27 \\ \Rightarrow 90-9\mathrm{x}=9\mathrm{x}+36 \\ \Rightarrow 9\mathrm{x}+9\mathrm{x}=90-36 \\ \Rightarrow 18\mathrm{x}=54 \\ \Rightarrow \mathrm{x}=\frac{54}{18} \\ \Rightarrow \mathrm{x}=3 \end{gathered}$$

Therefore the digit at ten's place is $3$.

The digit at one's place $=9-3=6$

Then the number $=10\times 3+6=30+6=36$

Therefore the required number is $36$.