There are 'roughly' 20 numbers in every 100 that are divisible by 5. There are 'roughly' 900 three digit numbers. So there are 'roughly' 180 three digit numbers divisible by 5. That's enough off from 136 to make me agree with Defennder and Tedjin. You must be wrong.
a_{n}=nth term of an AP (Arithmetic progression)
a=1st term
d=common difference
Since the 1st 3 digit no. divisible by 5 is 100
a=100
the last 3 digit no. divisible by 5 is 995
a_{n}=995
common difference (d)=5
n=no. of 3 digit nos. divisible by 5