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?

Explanation

let tens digite of number is`x`

let unit digit be`y`

number= `(10x+y)`

if no is reversed then

unit digit `x` & tens digit `y`

reversed number= `(10y+x)`

acc to question

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

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