If the number 357*25* is divisible by both 3 and 5, then the missing digits in the unit's place and the thousandth's place respectively are :

Answer: Option D

Let the unit's place be x and the thousand's place be y.
Then, 357y25x is divisible by 5 only when x = 0 or x = 5
Also, this number is divisible by 3 only when sum of its digits is divisible by 3
So, (22 + x + y) must be divisible by 3
∴ x + y = 2
Taking x = 0, we get y = 2
So, the unit place = 0 and thousand's place = 2