HCF of 4 times 27 times 3125,8 times 9 times | vedclass

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Question

(A) First,express each number in its prime factorization form:$4 	imes 27 	imes 3125 = 2^2 	imes 3^3 	imes 5^5$$8 	imes 9 	imes 25 	imes 7 = 2^

Vedclass · Accepted Answer

$180$

Vedclass · Answer

(A) First,express each number in its prime factorization form:$4 	imes 27 	imes 3125 = 2^2 	imes 3^3 	imes 5^5$$8 	imes 9 	imes 25 	imes 7 = 2^3 	imes 3^2 	imes 5^2 	imes 7^1$$16 	imes 81 	imes 5 	imes 11 	imes 49 = 2^4 	imes 3^4 	imes 5^1 	imes 11^1 	imes 7^2$To find the $HCF$,take the lowest power of each common prime factor present in all three numbers:Common prime factors are $2, 3,$ and $5$.Lowest power of $2$ is $2^2 = 4$.Lowest power of $3$ is $3^2 = 9$.Lowest power of $5$ is $5^1 = 5$.Therefore,$HCF = 2^2 	imes 3^2 	imes 5 = 4 	imes 9 	imes 5 = 180$.

$HCF$ of $4 \times 27 \times 3125$,$8 \times 9 \times 25 \times 7$ and $16 \times 81 \times 5 \times 11 \times 49$ is

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