# Find L.C.M. an H.C.F. of $72$ and $108$ by prime factorization method.

Last updated date: 21st Mar 2023

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Answer

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Hint: Here L.C.M. is the smallest positive integer that is divisible by each of them and H.C.F. is the largest positive integer that divides each of the integers.

Complete step-by-step answer:

Applying prime factorization method to $72$ we have,

$

72 = {2^3} \times {3^2} \\

72 = 2 \times 2 \times 2 \times 3 \times 3 \\

$

Applying prime factorization method to $108$ we have,

$

108 = {2^2} \times {3^3} \\

108 = 2 \times 2 \times 3 \times 3 \times 3 \\

$

$72$ is a product of three $2s$ and two $3s$

$108$ is a product of two $2s$ and three $3s$.

So, in this case the common factors are two $2s$ and two $3s$ while extra factors are one $2$ of $72$ and one $3$ of $108$.

So,

L.C.M. of $72$ and $108$\[ = 2 \times 2 \times 3 \times 3 \times 2 \times 3 = 216\]

H.C.F. of $72$ and $108$\[ = 2 \times 2 \times 3 \times 3 = 36\]

Thus, L.C.M and H.C.F of $72$ and $108$ are \[216\] and \[36\] respectively.

Note: In these types of problems for checking our answer we can use the relationship \[{\text{product of two numbers = }}\left( {{\text{H}}{\text{.C}}{\text{.F}}} \right)\left({{\text{L}}{\text{.C}}{\text{.M}}} \right)\].

Complete step-by-step answer:

Applying prime factorization method to $72$ we have,

$

72 = {2^3} \times {3^2} \\

72 = 2 \times 2 \times 2 \times 3 \times 3 \\

$

Applying prime factorization method to $108$ we have,

$

108 = {2^2} \times {3^3} \\

108 = 2 \times 2 \times 3 \times 3 \times 3 \\

$

$72$ is a product of three $2s$ and two $3s$

$108$ is a product of two $2s$ and three $3s$.

So, in this case the common factors are two $2s$ and two $3s$ while extra factors are one $2$ of $72$ and one $3$ of $108$.

So,

L.C.M. of $72$ and $108$\[ = 2 \times 2 \times 3 \times 3 \times 2 \times 3 = 216\]

H.C.F. of $72$ and $108$\[ = 2 \times 2 \times 3 \times 3 = 36\]

Thus, L.C.M and H.C.F of $72$ and $108$ are \[216\] and \[36\] respectively.

Note: In these types of problems for checking our answer we can use the relationship \[{\text{product of two numbers = }}\left( {{\text{H}}{\text{.C}}{\text{.F}}} \right)\left({{\text{L}}{\text{.C}}{\text{.M}}} \right)\].

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