The smallest natural number which is divisible by 18, 5, 81 and 90 is:

Correct option is A

Given:

Smallest natural number divisible by 18, 5, 81, and 90:

Solution:

Prime Factorization: Find the prime factorization of each number:

18 = 2 $$\times$$​ $$3^2$$​​

5 = 5

81 =$$3^4$$​​

90 = $$2 \times 3^2 \times 5$$​​

LCM: Multiply the highest powers of the prime factors together:

LCM = $$2^1 \times 3^4\times 5^1 = 2\times 81 \times 5 = 810$$​​

Therefore, the smallest natural number divisible by 18, 5, 81, and 90 is 810.

Option (A) is right.