What least number should be added to 6,000 to get a number exactly divisible by 19?

1. 9

2. 8

3. 4

4. 6

Given number = 6,000.

$\begin{array}{l} \phantom{x^21}{\quad 315} \ 19\overline{\smash{\big)}\quad 6000} \ \phantom{x^ + )}\phantom{)}\underline{-57} \ \phantom{{x^2 } + 2a} 30 \ \phantom{{x} +1a}\underline{-19} \ \phantom{{x^2 } + 2xa } 110 \ \phantom{{x} +1xa}\underline{-95} \ \phantom{{x^2 + 3x +)}} 15 \ \end{array}$

The remainder when 6000 is divided by 19 is 15.

Least number to be added to make the number divisible = 19 − 15 = 4.

Hence, option 3 is the correct option.