How many three digit numbers are there which are divisible by 5
[a] 180
[b] 190
[c] 150
[d] 170

- Hint: Use the fundamental principle of counting to find the number of three digit numbers divisible by 5. Use the property that if a number is divisible by 5, then there are 0 or 5 at unit place.

Complete step-by-step solution -

The fundamental principle of counting: According to the fundamental theorem of counting, if a task can be done in n ways and another task can be done in m ways, then the number of ways in which both the tasks can be done is mn, and the number of ways in which either of the tasks can be done is m+n.
In the case of forming numbers, we have to fill the decimal places. When we create a three digit number, we need to fill the units places, the tens place and the hundreds place.
Filling the hundreds place: The hundreds place cannot be filled by 0. Hence the number of ways in which units can be filled is 9.
Filling the units place: The units place can be filled with only two digits 0 or 5. This is because a number is divisible by 5 if and only if the digit at units place is 0 or 5.
Filling the tens place: The tens place can be filled in 10 ways.
Hence the total number of three digit numbers divisible by 5 is $9\times 2\times 10=180$.
Hence there are 180 three digit numbers divisible by 5.
Hence option [a] is correct.

Note: Alternative Solution:
We know that the number of numbers from 1-n divisible by k is $\left[ \dfrac{n}{k} \right]$, where [.] denotes the greatest integer function.
Hence the number of numbers from 1-99 divisible by 5 $=\left[ \dfrac{99}{5} \right]=19$
The number of numbers from 1-999 divisible by 5 $=\left[ \dfrac{999}{5} \right]=199$
Hence the number of three digit numbers divisible by 5 is $199-19=180$, which is the same as obtained above.