Find the greatest 3-digit number which is exactly divisible by 8, 20 and 24.

First, we find the LCM of 8, 20 and 24.

LCM of 8, 20 and 24 = 2 × 2 × 2 × 3 × 5 = 120.

The greatest 3-digit number is 999.

We divide 999 by 120 and find the remainder.

$\begin{array}{l} \phantom{x^2 1}{\quad 8} \ 120\overline{\smash{\big)}999} \ \phantom{x^ + }\phantom{}\underline{-960} \ \phantom{{x^2 } + (} 39 \ \end{array}$

The remainder is 39.

The required greatest 3-digit number = 999 − 39 = 960.

Hence, the greatest 3-digit number divisible by 8, 20 and 24 is 960.