Question

Use a method of your own choice to find the H.C.F. of the following:
(iv) $24,36,60\text{and}132$

Answer 1

(iv) Factor of a number: The number by which the given number can be divided completely
Prime factors of $24=2\times 2\times 2\times 3$
Prime factors of $36=2\times 2\times 3\times 3$
Prime factors of $60=2\times 2\times 3\times 5$
Prime factors of $132=2\times 2\times 3\times 11$
$\therefore$ Common prime factors of 24, 36, 60 and $132=2\times 2\times 3$
Hence, H.C.F. of 24, 36, 60 and $132=2\times 2\times 3=12$