Find the HCF of the given numbers by division method:

(i) 198, 429

(ii) 20, 64, 104

(iii) 120, 144, 204

(i) 198, 429

$\begin{array}{l} 198\overline{\smash{\big)}429\smash{\big(}}2\phantom{)} \ \phantom{x}\phantom{))}\underline{-396} \ \phantom{{x^2 } (()))}33\overline{\smash{\big)}198\smash{\big(}}\phantom{)}6 \ \phantom{{x} +1xa)}\underline{-198} \ \phantom{{x^2 a} + 2x+()} 0 \ \end{array}$

Hence, HCF of 198 and 429 = 33.

(ii) 20, 64, 104

First, find HCF of 20 and 64:

$\begin{array}{l} 20\overline{\smash{\big)}64\smash{\big(}}\phantom{)}3 \ \phantom{x}\phantom{}\underline{-60} \ \phantom{{x^2 } 1)}4\overline{\smash{\big)}20\smash{\big(}}\phantom{}5 \ \phantom{{x} +(}\underline{-20} \ \phantom{{x^2 a} + 2} 0 \ \end{array}$

So, HCF of 20 and 64 = 4.

Now, find HCF of 4 and 104:

$\begin{array}{l} 4\overline{\smash{\big)}104\smash{\big(}}\phantom{)}26 \ \phantom{(}\phantom{}\underline{-104} \ \phantom{{x^2 } +)}0 \ \end{array}$

So, HCF of 4 and 104 = 4.

Hence, HCF of 20, 64 and 104 = 4.

(iii) 120, 144, 204

First, find HCF of 120 and 144:

$\begin{array}{l} 120\overline{\smash{\big)}144\smash{\big(}}\phantom{}1 \ \phantom{x}\phantom{))}\underline{-120} \ \phantom{{x^2 } ()1)}24\overline{\smash{\big)}120\smash{\big(}}\phantom{}5 \ \phantom{{x} +1xa}\underline{-120} \ \phantom{{x^2 a} + 2x+} 0 \ \end{array}$

So, HCF of 120 and 144 = 24.

Now, find HCF of 24 and 204:

$\begin{array}{l} 24\overline{\smash{\big)}204\smash{\big(}}\phantom{}8 \ \phantom{x}\phantom{}\underline{-192} \ \phantom{{x^2 } 1)}12\overline{\smash{\big)}24\smash{\big(}}\phantom{}2 \ \phantom{{x} +1(}\underline{-24} \ \phantom{{x^2 a} + 2x} 0 \ \end{array}$

So, HCF of 24 and 204 = 12.

Hence, HCF of 120, 144 and 204 = 12.