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.

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Updated on: 10-Oct-2022

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