What is the smallest number which on subtracting 9 from it becomes divisible by 24, 33, 42 and 56?

Solution:

We need the smallest number such that
(number − 9) is divisible by 24, 33, 42, and 56.

Step 1: Find LCM of 24, 33, 42, 56

Prime factors:

24 = $2^3 \times 3$
33 = $3 \times 11$
42 = $2 \times 3 \times 7$
56 = $2^3 \times 7$

LCM = $2^3 \times 3 \times 7 \times 11 = 1848$

Step 2: Required number

Number = $1848 + 9 = 1857$

✅ Final Answer: 1,857