# Possible values of x and y?

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)

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