In a two-digit natural number, the digit at th etens place is equal to the square of the digit at units place. If 54 is subtracted from the number, the digits get interchanged . Find the number.

In a two-digit natural number, the digit at th etens place is equal to

1.	In a two-digit natural number, the digit at th etens place is equal to the square of the digit at units place. If 54 is subtracted from the number, the digits get interchanged . Find the number.
Answer» Correct Answer - The required natural number is 93 Let the digit at the units place be x. Then the digit at the tens place is `x^(2)`. The number is `10x^(2) + x ` . The number obtained by interchanging the digits is `10x + x^(2)`. From the given condition, `10x^(2) +x- 54 = 10 x + x^(2) ` `: 10x^(2) - x^(2) + x- 10 x - 54 = 0 ` ` :. 9 x^(2) - 9x - 54 =0 ` `:. x^(2) - x -6 = 0 ` .....(Dividing by 9 ) `:. x^(2)-3x + 2 x - 6 = 0` `:. x( x-3) + 2 (x-3) = 0 ` `:. ( x-3) ( x+ 2) = 0 ` `:. x-3 =0` or ` x+ 2 = 0 ` `:. x = 3 ` or `x = - 2 ` But the natural cannot negative `:. x = - 2 ` is unacceptable `:. x = 3 ` and `x^(2) = ( 3)^(2) =9 ` The number is `10x^(2) + x = 10 ( 9) + 3 = 90 + 3 = 93 `