Answer this question.

Let's solve this problem without using any mathematical formula of Arithmetic Progression. This gives us the speedy answer which is so required in the competitive exams where there is a constraint of time.
The greatest three digit number is 999.
And 999/19 = 52.57.
This implies upto 999, a total of 52 numbers are completely divisible.
Subtracting from this, number of two- digit numbers also divisible by 19 taken into account in the previous calculation, i.e 19,38,57,76, and 95, we get :
52-5 = 47 
Thus there are 47 such 3 digit numbers in all.