Find the smallest 4-digit number which is exactly divisible by 32, 36 and 48.

First, we find the LCM of 32, 36 and 48.

LCM of 32, 36 and 48 = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288.

The smallest 4-digit number is 1000.

We divide 1000 by 288 and find the remainder.

$\begin{array}{l} \phantom{x^2 1}{\quad 3} \ 288\overline{\smash{\big)}1000} \ \phantom{x^ + )}\phantom{)}\underline{-864} \ \phantom{{x^2 } + 2} 136 \ \end{array}$

The remainder is 136.

The required smallest 4-digit number = 1000 + (288 − 136) = 1000 + 152 = 1152.

Hence, the smallest 4-digit number divisible by 32, 36 and 48 is 1152.