Find the HCF of the given numbers by division method:

(i) 54, 82

(ii) 84, 120, 156

(i) 54, 82

$\begin{array}{l} 54\overline{\smash{\big)}82\smash{\big(}}\phantom{}1 \ \phantom{x}\phantom{}\underline{-54} \ \phantom{{x^2 } 1}28\overline{\smash{\big)}54\smash{\big(}}\phantom{}1 \ \phantom{{x} +1}\underline{-28} \ \phantom{{x^2 a} + 2} 26\overline{\smash{\big)}28\smash{\big(}}\phantom{}1 \ \phantom{{x} +1xa()}\underline{-26} \ \phantom{{x^2 a} + 2x+(} 2\overline{\smash{\big)}26\smash{\big(}}\phantom{}13 \ \phantom{{x} +1xa++(}\underline{-26} \ \phantom{{x^2 a} + 2x+++} 0 \ \end{array}$

Hence, HCF of 54 and 82 = 2.

(ii) 84, 120, 156

First, find HCF of 84 and 120:

$\begin{array}{l} 84\overline{\smash{\big)}120\smash{\big(}}\phantom{}1 \ \phantom{x}\phantom{)}\underline{-84} \ \phantom{{x^2 } 1)}36\overline{\smash{\big)}84\smash{\big(}}\phantom{}2 \ \phantom{{x} +1(}\underline{-72} \ \phantom{{x^2 a} + 2(} 12\overline{\smash{\big)}36\smash{\big(}}\phantom{}3 \ \phantom{{x} +1xa()}\underline{-36} \ \phantom{{x^2 a} + 2x+(} 0 \ \end{array}$

So, HCF of 84 and 120 = 12.

Now, find HCF of 12 and 156:

$\begin{array}{l} 12\overline{\smash{\big)}156\smash{\big(}}\phantom{}13 \ \phantom{x}\phantom{)}\underline{-156} \ \phantom{{x^2 } 1+)}0 \ \end{array}$

So, HCF of 12 and 156 = 12.

Hence, HCF of 84, 120 and 156 = 12.