The smallest natural number which is divisible by 9, 6, 72 and 76 is:

Correct option is B

Given:
We need to find the smallest natural number divisible by 9, 6, 72, and 76.
Concept Used:
The smallest natural number divisible by a set of numbers is the Least Common Multiple (LCM) of those numbers.
Solution:
Find the prime factorization of each number:

9 = $$3^2$$​​

6 = $$2 \times 3$$​​

72 = $$2^3 \times 3^2$$​

76 = $$2^2 \times 19$$​ 

$$\text{LCM} = 2^3 \times 3^2 \times 19$$​​
$$\text{LCM} = 8 \times 9 \times 19$$​​
$$\text{LCM} = 72 \times 19 = 1,368$$​
Thus, option (b) is right.