Answer :

Let x, y and z be the digits in a three-digit number A = xyz.


Reversing the digits, we get B = zyx.


Sum of the digits in A is (100x + 10y + z).


Sum of the digits in B is (100z + 10y + x).


Difference S = A + B = (100x + 10y + z) - (100z + 10y + x).


S = 100x-x + 10y-10y + z-100z = 99x-99z


99(x-z) = 9x11(x-z)


The number is divisible by 9 and 11.


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