If the 6-digit number 608xy0 is divisible by both 3 and 11, then the non-zero digits in the hundred's place (x) and ten's place (y), respectively, are:

Given the number is 608xy0. It is divisible by both 3 and 11.
Divisibility rule of 11:
The difference between the sum of digits at odd places and the sum of digits at even places must be 0 or a multiple of 11.

Sum of odd positions (from right)=0+x+0=x
Sum of even positions (from right)=y+8+6=y+14
The difference:

Difference=(y+14)−x=y−x+14
For divisibility by 11, this value can be 11 or 22 (since x and y are single digits, y−x+14=0⟹x−y=14, which is impossible).

Case 1: y−x+14=11⟹x−y=3
Case 2: y−x+14=22⟹y−x=8⟹x−y=−8

Divisibility rule of 3:
The sum of all digits must be divisible by 3.

Sum of digits=6+0+8+x+y+0=14+x+y
Let us check the given options:

Option A: x=6,y=5

Check x−y=6−5=1

=3 or −8. (Incorrect)

Option B: x=5,y=6

Check x−y=5−6=−1

=3 or −8. (Incorrect)

Option C: x=5,y=8

Check x−y=5−8=−3

=3 or −8. (Incorrect)

Option D: x=8,y=5

Check x−y=8−5=3 (Satisfies Case 1).

Let's check the sum of digits for x=8,y=5:

Sum=14+8+5=27
Since 27 is divisible by 3, this option is completely correct.