Possible values of x and y?

Solution:

If the sum of digits of the number is divisible by 9, then the number itself is divisible by 9.
The given number is 6x321y
Sum of the digits = 6 + x + 3 + 2 + 1 + y
= 12 + x + y should be divisible by 9

For it to be divisible by 9 we will have two cases: x + y = 6 and x + y = 15
For x + y = 6
If x = 0, y = 6
x = 1, y = 5
x = 2 , y = 4
x = 3 , y = 3
x = 4 , y = 2
x = 5 , y = 1
x = 6, y = 0

For x + y = 15 [ x and y should be single digits]
If x = 6, y = 9
x = 7, y = 8
x = 8, y = 7
x = 9, y = 6
​​​​​​
Hence, the total no.of the possible value of x and y is 11.

Therefore Option (b)