If the square of a two-digit number is reduced by the square of the number formed by reversing the digits of the number, the final result is :

A. Divisible by 11

B. Divisible by 9

C. Necessarily irrational

D. Both (A) and (B)

Answer: Option D

Solution(By Examveda Team)

Let the two-digit number be 10x + y
Then, number formed by reversing the digits = 10y + x
Difference of square of the numbers :
$$ = {\left( {10x + y} \right)^2} - {\left( {10y + x} \right)^2}$$
$$ = \left( {100{x^2} + {y^2} + 20xy} \right) - $$     $$\left( {100{y^2} + {x^2} + 20xy} \right)$$
$$ = 99\left( {{x^2} - {y^2}} \right)$$     which is divisible by both 9 and 11