Z Given, number 8475639AB is divisible by 99 , then it also divisible by 9 and 11. A number is divisible by 9 , if the sum of the digit of that number is divisible by 9. Now, 8+4+7+5+6+3+9+A+B =42+A+B Here, the possible value of A+B=3 and 12 Now, a number is divisible by 11 if the difference of the sum of the even and odd places digits is equal to zero or the multiple of 11 . Then, (8+7+6+9+B) −(4+5+3+A) =30+B−12−A =18+B−A Here, the possible value of B−A=4 So, the value of A+B is can not be equal to 3 . Hence, A=4,B=8