The smallest natural number which is divisible by 48, 72, 16 and 15 is:

Correct option is B

Given:

The problem asks to find the smallest natural number divisible by 48, 72, 16, and 15.

Formula Used:

LCM of numbers = Product of the highest powers of prime factors of all the numbers.

Solution:

Prime factorization of each number:

48 = $$2^4 \times 3$$​​

72 = $$2^3 \times 3^2$$​​

16 = $$2^4$$​ 

15 = $$3 \times 5$$​​

LCM = $$2^4 \times 3^2 \times 5 = 16 \times 9 \times 5$$​ = 720

Thus, the smallest natural number divisible by 48, 72, 16, and 15 is 720.