Question

The sum of a two-digit number and the number obtained by reversing the order of its digits
is 121, and the two digits differ by 3. Find the number.

Answer 1

Let x be the ones digit and y be the tens digit. Then
Two digit number before reversing = 10y + x
Two digit number after reversing = 10x + y
As per the question
$$\begin{gathered} \left(10y+x\right)+\left(10x+y\right)=121 \\ \Rightarrow 11x+11y=121 \\ \Rightarrow x+y=11.....\left(i\right) \end{gathered}$$
Since the digits differ by 3, so
$x-y=3......\left(ii\right)$
Adding (i) and (ii), we get
$2x=14\Rightarrow x=7$
Putting x = 7 in (i), we get
$7+y=11\Rightarrow y=4$
Changing the role of x and y, x = 4 and y = 7
Hence, the two-digit number is 74 or 47.