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If HCF of \[\left( {26,169} \right) = 13\]. Then LCM of \[\left( {26,169} \right)\] is
A) 26
B) 52
C) 338
D) 13

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Last updated date: 20th Jun 2024
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Answer
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Hint:
Here, we will use the relation between H.C.F. and L.C.M. to find the L.C.M. of two numbers. The Highest Common Factor (H.C.F) of two numbers is defined as the greatest number which divides exactly both the numbers. The Least Common Multiple (L.C.M) of two numbers is defined as the smallest number which is divisible by both the numbers.

Complete step by step solution:
It is given that the highest common factor (H.C.F) of the numbers 26, 169 as 13.
Now, we will find the least common multiple of the numbers 26, 169.
We know that the product of the least common multiple and the highest common factor of the natural numbers is always equal to the product of the natural numbers. Mathematically,
L.C.M. of the given numbers \[ \times \] H.C.F. of the given numbers \[ = \] Product of the given numbers
Let the L.C.M of \[\left( {26,169} \right)\] be \[x\].
By substituting H.C.F. of the number as 13 in the above formula, we get
 \[ \Rightarrow x \times 13 = \left( {26} \right) \times \left( {169} \right)\]
Dividing both sides by 13, we get
 \[ \Rightarrow x = \dfrac{{\left( {26} \right) \times \left( {169} \right)}}{{13}}\]
Dividing 26 by 13, we get
 \[ \Rightarrow x = \left( 2 \right) \times \left( {169} \right)\]
Multiplying the terms, we get
 \[ \Rightarrow x = 338\]
$\therefore $ L.C.M. of the given numbers \[\left( {26,169} \right)\] \[ = 338\]
Therefore, the L.C.M. of \[\left( {26,169} \right)\] is 338.

Thus Option (C) is the correct answer.

Note:
We can find the least common multiple of the numbers by just doing the prime factorization of the numbers. Since we are provided with the highest common factor, we are using the relation between two numbers. Also, we are provided with two co-prime numbers. We know that the H.C.F. of two co-prime numbers is 1. Thus, the least common multiple of two co-prime numbers is equal to the product of the given two co-prime numbers.