Question

find LCM and HCF of following pair of integer and verify that LCM × HCF = product of two numbers


(1. )26 and 91​

Answer 1

If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs. So the product of L.C.M and H.C.F is (182×13)=2366. And the product of two numbers is (26×91)=2366.

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Answer 2

$\huge\underline\mathsf\purple{Answer:- }$

$\underline{\frak{\dag \; lcm \: of \: \; 26: - }}$

$\begin{gathered}\Large{ \begin{array}{c|c|c} \tt 2 & \sf \orange{26 }& \sf \\ \tt 13& \sf \orange{13}& \orange{ }  \\ & \sf \orange{1 } & \sf \orange{ }\end{array}}\end{gathered}$

$\underline{\frak{\dag \; lcm \: of \; 91: - }}$

$\begin{gathered}\Large{ \begin{array}{c|c|c} \tt 7 & \sf \orange{91}& \sf \\ \tt 13& \sf \orange{13}& \orange{ }  \\ & \sf \orange{1 } & \sf \orange{ }\end{array}}\end{gathered}$

$\begin{gathered} \sf \purple{ 26 = 2 \times 13} \: \\ \\ \sf \pink{ 91= 7 \times 13} \\ \\ \sf \green{hcf= 13} \\ \\ \sf \red{ \boxed{ \bf{lcm= 2 \times 7 \times 13 = 182}}}\end{gathered}$

$\boxed{ \red{ \sf product \: of \: two \: number = 26 \times 91 = 2366}} \:$

$\boxed{ \red{\sf hcf \times lcm = 13 \times 182 = 2366}} \:$

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