The sum of a two–digit number and the number obtained by reversing the digits is 66. If the digits of the number 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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