Question

A number consists of two digits whose sum is five. When the digits are reversed the number becomes greater by nine. Find the number.

Answer 1

We have to find a two-digit number whose sum of digits is $5$ and when the digits are reversed the number becomes greater by $9$.

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

Then the digit at one's place $=5-\mathrm{x}$ (since the sum of digits is $5$)

Thus the number $=10\mathrm{x}+\left(5-\mathrm{x}\right)$

The number formed by reversing the digits $=10\left(5-\mathrm{x}\right)+\mathrm{x}$

According to the question,

$$\begin{gathered} 10\left(5-\mathrm{x}\right)+\mathrm{x}=10\mathrm{x}+\left(5-\mathrm{x}\right)+9 \\ \Rightarrow 50-10\mathrm{x}+\mathrm{x}=10\mathrm{x}+5-\mathrm{x}+9 \\ \Rightarrow 50-9\mathrm{x}=9\mathrm{x}+14 \\ \Rightarrow 9\mathrm{x}+9\mathrm{x}=50-14 \\ \Rightarrow 18\mathrm{x}=36 \\ \Rightarrow \mathrm{x}=\frac{36}{18} \\ \Rightarrow \mathrm{x}=2 \end{gathered}$$

Therefore the digit at ten's place is $2$.

The digit at one's place $=5-2=3$

Then the number $=10\times 2+3=20+3=23$

Therefore the required number is $23$.