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If the L.C.M of two numbers such that $L.C.M\left( {91,26} \right) = 182$, then find the H.C.F of (91, 26).

Answer
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Hint – In this question use the concept that the product of two numbers is equal to the product of L.C.M and the H.C.F. The two numbers are 91 and 26. So simple substitution of values to this concept will get the H.C.F of required numbers.

Complete step by step answer:
As we know that the H.C.F of the numbers is the product of common factors.
Least common multiple of two numbers is to first list the prime factors of each number. Then multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs.
The L.C.M of 26 and 91 is 182
$ \Rightarrow L.C.M\left( {91,26} \right) = 182$ (Given).
And the product of two numbers is $\left( {26 \times 91} \right) = 2366$.
Now as we know that L.C.M $ \times $ H.C.F = product of two numbers.
Therefore, H.C.F = $\dfrac{{{\text{product of two numbers}}}}{{L.C.M}}$
$ \Rightarrow H.C.F\left( {91,26} \right) = \dfrac{{2366}}{{182}} = 13$
So the required H.C.F of 91 and 26 is 13.
So, this is the required answer.

Note – The least common multiple or the lowest common multiple of any two integers is the smallest positive integer that is divisible by both the numbers. The H.C.F or the greatest common factor of any two numbers is the largest factor which is common to both the numbers.