Using Prime Factorisation Method, determine the HCF of the numbers : $40 a n d 48$

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Solution

First of all lets find the prime factors of both $40, 48$
$2$ $40$ $2$ $48$
$2$ $20$ $2$ $24$
$2$ $10$ $2$ $12$
$5$ $5$ $2$ $6$
$1$ $3$ $3$
$1$
Prime factors of $40$ are $2 \times 2 \times 2 \times 5$ Prime factors of $48$ are $2 \times 2 \times 2 \times 2 \times 3$
Now to find the HCF we need to common factors of both the numbers
which is $2$ (which occurs three time in both)
so $H C F$ of $40, 48 i s = 2 \times 2 \times 2 = 8$

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$2$	$40$	$2$	$48$
$2$	$20$	$2$	$24$
$2$	$10$	$2$	$12$
$5$	$5$	$2$	$6$
	$1$	$3$	$3$
			$1$
Prime factors of $40$ are $2 \times 2 \times 2 \times 5$		Prime factors of $48$ are $2 \times 2 \times 2 \times 2 \times 3$