Find the HCF of 96 and 404 by the prime facto | vedclass

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Question

(A) Step $1$: Find the prime factorization of $96$ and $404$.$96 = 2^5 	imes 3^1$$404 = 2^2 	imes 101^1$Step $2$: Find the $HCF$ by taking the small

Vedclass · Accepted Answer

$HCF = 4, LCM = 9696$

Vedclass · Answer

(A) Step $1$: Find the prime factorization of $96$ and $404$.$96 = 2^5 	imes 3^1$$404 = 2^2 	imes 101^1$Step $2$: Find the $HCF$ by taking the smallest power of each common prime factor.$HCF(96, 404) = 2^2 = 4$.Step $3$: Use the relationship $LCM(a, b) 	imes HCF(a, b) = a 	imes b$.$LCM(96, 404) = \frac{96 	imes 404}{HCF(96, 404)}$$LCM(96, 404) = \frac{96 	imes 404}{4} = 96 	imes 101 = 9696$.

$Q.1.$ $HCF$ of $25$ and $10$	$A. 5$
$Q.2.$ $LCM$ of $17$ and $11$	$B. 187$

Find the $HCF$ of $96$ and $404$ by the prime factorization method,and hence find their $LCM$.

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For any two positive integers $a$ and $b$,what is the relationship between their $HCF$ and $LCM$? $\text{HCF}(a, b) \times \text{LCM}(a, b) = \text{?}$

The smallest positive number greater than $5$ such that when it is divided by $20, 30,$ or $40,$ it leaves the remainder $5$ is:

Find the $HCF$ of $12$,$15$,and $21$.

Match the following items related to the properties of numbers:
$Q.1.$ $HCF$ of $25$ and $10$ $A. 5$
$Q.2.$ $LCM$ of $17$ and $11$ $B. 187$

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