How many numbers divisible by 125 can be written by using all the digits 2,3,8,7,5 exactly once?
A.4
B.5
C.6
D.7

Hint: In this type of question remember the divisibility rule of the numbers like $2,3,5,7,11,9$
Using this we can find the divisibility of factor of any numbers.

Complete Step-by-step Solution
Step: given number is $125$
By divisibility rule
We know that any numbers whose last digits are a multiple of $125$ is divisible by $125$
The only $2$ such numbers for the last $3$ digits are \[375{\text{ }}and{\text{ }}875\] leaving $2$ digits that can be pretended in 2 ways each
For units place, we have only one number
For tens place, we have only one number
For a hundred places, we have only two numbers.
For the remaining two places we have two numbers.
Therefore we have $\left( {1 \times 1 \times 2 \times 2} \right) = 4$
So there exactly 4 numbers they are:
$
  28375 \\
  82375 \\
  23875 \\
  32875 \\
$
$\therefore $ Four numbers divisible by 125 can be written by using all the digits 2,3,8,7,5 exactly once.

Additional information:
A Divisibility rule is a way of determining whether a given number or integer is divisible by a fixed divisor without performing division. For example, 2 is divisible by 4,6,8…

Note:
As we know for a number to be divisible by 125, its last three digits should be divisible by 125.
Similarly, all divisibility rules are applicable to other factors of numbers accordingly. Using the divisibility rule of $5$, We can say as 125 is divisible by 5 as the last digit is divisible by 5 then complete number 125 is divisible by $5$