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The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.
Given :
The sum of a two-digit number and the number obtained by reversing the order of its digits is 99.
The two digits differ by 3.
To do :
We have to find the original number.
Solution :
Let the two-digit number be $10x+y$.
$x-y=3$ or $y-x=3$
If the digits are interchanged and the resulting number is added to the original number, we get 99.
This implies,
$(10x+y)+(10y+x) = 99$
$10x+x+10y+y=99$
$11x+11y=99$
$11(x+y)=99$
$x+y=\frac{99}{11}$
$x+y=9$
If $x-y=3$ and $x+y=9$
$x-y+x+y=3+9$
$2x=12$
$x=6$
This implies,
$6+y=9$
$y=9-6=3$
$x = 6, y=3$
Then the original number is $10x+y=10(6)+3=63$
If $y-x=3$ and $x+y=9$
$y-x+x+y=3+9$
$2y=12$
$y=\frac{12}{2}$
$y=6$
This implies,
$x+6=9$
$x=9-6=3$
Then the original number is $10x+y=10(3)+6=36$
Therefore, the original number is 63 or 36.