Question

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

Answer 1

Let the digit in the unit's place be $x$ and the digit in the ten's place be $y$ . Then,

Number$=10y+x$

The number obtained by interchanging the digits is $10x+y$

According to the given conditions,

$(10y+x)+(10x+y)=99$

$=11x+11y=99$

$=x+y=9$ ......1

And,$x-y=\pm 3$ ......2 (given)

On solving equation 1 and 2 ,we get

$x=6,y=3orx=3,y=6$

Hence the required number is either$63or36$