Question

Find the HCF of 126, 162 and 180 using prime factorization method for HCF. Step by Step method and also solve and share the photo of it plss

Answer 1

$\large\underline{\sf{Solution-}}$

Consider

$\red{\rm :\longmapsto\:Prime \: factorization \: of \: 126}$

$\red{\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:126 \:\:}}}\\ {\underline{\sf{3}}}& \underline{\sf{\:\:63 \:\:}} \\\underline{\sf{3}}&\underline{\sf{\:\:21\:\:}} \\ {\underline{\sf{7}}}& \underline{\sf{\:\:7 \:\:}} \\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}}$

Thus,

$\red{\rm :\longmapsto\:Prime \: factorization \: of \: 126 = 2 \times {3}^{2} \times 7 }$

Consider

$\green{\rm :\longmapsto\:Prime \: factorization \: of \: 162}$

$\green{\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:162 \:\:}}}\\ {\underline{\sf{3}}}& \underline{\sf{\:\:81 \:\:}} \\\underline{\sf{3}}&\underline{\sf{\:\:27\:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:9 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:3\:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}}$

Thus,

$\green{\rm :\longmapsto\:Prime \: factorization \: of \: 162 = 2 \times {3}^{4} }$

Consider

$\blue{\rm :\longmapsto\:Prime \: factorization \: of \: 180}$

$\blue{\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:180 \:\:}}}\\ {\underline{\sf{3}}}& \underline{\sf{\:\:90 \:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:30\:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:15 \:\:}} \\ {\underline{\sf{5}}}& \underline{\sf{\:\:5\:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}}$

Thus,

$\blue{\rm :\longmapsto\:Prime \: factorization \: of \: 180 = {2}^{2} \times {3}^{2} \times 5}$

So, we have

$\red{\rm :\longmapsto\:Prime \: factorization \: of \: 126 = 2 \times {3}^{2} \times 7 }$

$\green{\rm :\longmapsto\:Prime \: factorization \: of \: 162 = 2 \times {3}^{4} }$

$\blue{\rm :\longmapsto\:Prime \: factorization \: of \: 180 = {2}^{2} \times {3}^{2} \times 5}$

Hence,

$\purple{\rm :\longmapsto\: \boxed{ \bf{ \:HCF( 126, 162, 180 ) = {3}^{2} \times 2 = 18}}}$