Find the LCM and HCF of \(26 \)and \(91\) and verify that LCM \(\times\) HCF \(=\)Product of two numbers

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Find the LCM and HCF of \(26 \)and \(91\) and verify that LCM \(\times\) HCF \(=\)Product of two numbers

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Solution:

Doing prime factorization of \(26\) and \(91\).

\(26=2\times 13\) and \(91=7\times 13\)

\(\therefore \) LCM of \(26\) and \(91\) \(=2\times 7\times 13=182\)

and HCF of \(26\) and \(91 =13 \)

Now,

LCM \(\times \)HCF\(=182\times 13=2366\)

Product of two numbers\(=26\times 91=2366\)

Since,\(182\times 13=26\times 91\)

so, LCM \(\times\) HCF \(=\)Product of two numbers

Hence proved.

Doing prime factorization of \(26\) and \(91\).

\(26=2\times 13\) and \(91=7\times 13\)

\(\therefore \) LCM of \(26\) and \(91\) \(=2\times 7\times 13=182\)

and HCF of \(26\) and \(91 =13 \)

Now,

LCM \(\times \)HCF\(=182\times 13=2366\)

Product of two numbers\(=26\times 91=2366\)

Since,\(182\times 13=26\times 91\)

so, LCM \(\times\) HCF \(=\)Product of two numbers

Hence proved.

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