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Home » 2. Find the LCM and HCF of the following integers by applying the prime factorization method:(v)     (vi]    
2. Find the LCM and HCF of the following integers by applying the prime factorization method:(v)

    \[\mathbf{84},\text{ }\mathbf{90}\text{ }\mathbf{and}\text{ }\mathbf{120}\]

(vi]

    \[\mathbf{24},\text{ }\mathbf{15}\text{ }\mathbf{and}\text{ }\mathbf{36}\]

CBSE | Class 10 | Maths | RD Sharma | Real Numbers
2. Find the LCM and HCF of the following integers by applying the prime factorization method:(v)

    \[\mathbf{84},\text{ }\mathbf{90}\text{ }\mathbf{and}\text{ }\mathbf{120}\]

(vi]

    \[\mathbf{24},\text{ }\mathbf{15}\text{ }\mathbf{and}\text{ }\mathbf{36}\]

Solution:

First,

 Find the prime factors of the given integers:

    \[84,\text{ }90\text{ }and\text{ }120\]

     

For,

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> ~84\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }7 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> 90\text{ }=\text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }5 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> 120\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }5 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

Now,

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> L.C.M\text{ }of\text{ }84,\text{ }90\text{ }and\text{ }120\text{ }=\text{ }{{2}^{3}}~\times \text{ }{{3}^{2}}~\times \text{ }5\text{ }\times \text{ }7 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \therefore L.C.M\text{ }\left( 84,\text{ }90,\text{ }120 \right)\text{ }=\text{ }2520 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

And,

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> ~ \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> ~H.C.F\text{ }\left( 84,\text{ }90\text{ }and\text{ }120 \right)\text{ }=\text{ }6 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

Solution:

First,

find the prime factors of the given integers:

    \[24,\text{ }15\text{ }and\text{ }36\]

         

For,

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> 24\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }x\text{ }2\text{ }x~3 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> 15\text{ }=\text{ }3\text{ }\times \text{ }5 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> 36\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }3 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

Now,

    \[LCM\text{ }of\text{ }24,\text{ }15\text{ }and\text{ }36\text{ }=\text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }2\text{ }\times \text{ }3\text{ }\times \text{ }3\text{ }\times \text{ }5\text{ }=\text{ }{{2}^{3}}~x\text{ }{{3}^{2}}~x\text{ }5\]

∴

    \[LCM\text{ }\left( 24,\text{ }15,\text{ }36 \right)\text{ }=\text{ }360\]

And,    

    \[HCF\text{ }\left( 24,\text{ }15\text{ }and\text{ }36 \right)\text{ }=\text{ }3\]

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