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

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