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

Answer

Verified

383.1k+ views

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 \[{\text{product of two numbers = }}\left( {{\text{H}}{\text{.C}}{\text{.F}}} \right)\left({{\text{L}}{\text{.C}}{\text{.M}}} \right)\].

Recently Updated Pages

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which one of the following places is unlikely to be class 8 physics CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the past tense of read class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Elucidate the structure of fructose class 12 chemistry CBSE

What is pollution? How many types of pollution? Define it