Question

The sum of a two digit number and the number obtained by reversing the order of the digits is 99. Find the number, if the digits differ by 3.

Answer 1

Let the face value of the digits be x and y .
Now $xy=10\times x+y\times 1$ by using both place value and face value and in it's expanded form
e.g. $32=10\times 3+2\times 1$
When we reverse the number $xy$ it becomes $yx$ and again using both place value and face value and in it's expanded form the number becomes
$yx=10\times y+x\times 1$

Now..
According to the question

: x-y=3............................................ eq(1 )
$\to$ 10x+ y+ 10y+x = 99................................... eq (2)
So from eq (2) we have
$\to$10x+y+10y+x= 99
$\to$11x+11y=99
$\to$x+y=9 ...... eq (3)

Now
from eq (1)
we have
x-y=3
$\to$=3+y
substituing value of x in eq (3)
x+y=9
$\to$ 3+y+y=9
$\to$ 2y=6
$\to$ y=3
Substituting value of y in eq1
x-y=3
$\to$ x-3=3
$\to$ x=6
so the no. is 36 or 63