​$$\text{The HCF of } 2^{18} - 1 \text{ and } 2^{20} - 1 \text{ is } \_\_\_\_.$$​​

Correct option is B

Given:

We need HCF $$(2^{18} – 1 , 2^{20} – 1).$$​​

Formula Used:

For numbers of the form $$a^m - 1$$​ and $$a^n - 1:$$​​

​$$\text{HCF}(a^m - 1, \; a^n - 1) = a^{\gcd(m,n)} - 1$$​

Solution:

Here, a = 2, m = 18, n = 20.

gcd(18, 20) = 2

So,

​$$\text{HCF}(2^{18} - 1, \; 2^{20} - 1) = 2^{2} - 1 = 4 - 1 = 3$$​