The sum of a two-digit number and the number formed by reversing the order of digits is 66. If the two digits differ by 2, find the number. How many such numbers are there?

Given :

The sum of a two-digit number and the number formed by reversing the order of digits is 66.

The two digits differ by 2.

To do :

We have to find the original number.

Solution :

Let the two-digit number be $10x+y$.

$x-y=2$ or $y-x=2$

If the digits are interchanged and the resulting number is added to the original number, we get 66.

This implies,

$(10x+y)+(10y+x) = 66$

$10x+x+10y+y=66$

$11x+11y=66$

$11(x+y)=66$

$x+y=\frac{66}{11}$

$x+y=6$

If $x-y=2$ and $x+y=6$

$x-y+x+y=2+6$

$2x=8$

$x=4$

This implies,

$4+y=6$

$y=6-4=2$

$x = 4, y=2$

Then the original number is $10x+y=10(4)+2=42$

If $y-x=2$ and $x+y=6$

$y-x+x+y=2+6$

$2y=8$

$y=\frac{8}{2}$

$y=4$

This implies,

$x+4=6$

$x=6-4=2$

Then the original number is $10x+y=10(2)+4=24$

Therefore, the original number is 42 or 24. There are two numbers which satisfy the given conditions.

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Updated on: 10-Oct-2022

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