Find the HCF of 96 and 404 by prime factorisation method. Thereafter, find their L.C.M.

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Solution

$96 = (2)^{5} \times (3) 404 = (2)^{2} \times (101)$
H.C.F of these two integers
$= (2)^{2} = 4 L . C . M = \frac{(96) \times (404)}{4} = 9696$

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Find the HCF of $96$ and $404$ .

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Find the HCF of 96 and 404 by prime factorisation method. Thereafter, find their L.C.M.

96=(2)5×(3)404=(2)2×(101) H.C.F of these two integers =(2)2=4L.C.M=(96)×(404)4=9696

$96 = (2)^{5} \times (3) 404 = (2)^{2} \times (101)$
H.C.F of these two integers
$= (2)^{2} = 4 L . C . M = \frac{(96) \times (404)}{4} = 9696$