Question

LCM of 1125,144 2hat is the ans of the question​

Answer 1

Step-by-step explanation:

it never rains but it pours

Answer 2

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

Given numbers are

• 1125

• 144

➢ Let first find the prime factors of 144 and 1125.

Consider,

$\red{\rm :\longmapsto\:Prime \: factorization \: of \: 144}$

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:144 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:72 \:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:36\:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\: 18\:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:9\:\:}}\\\underline{\sf{3}}&\underline{\sf{\:\:3\:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

$\red{\rm :\longmapsto\:Prime \: factorization \: of \: 144 = {2}^{4} \times {3}^{2} }$

Consider

$\red{\rm :\longmapsto\:Prime \: factorization \: of \: 1125}$

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{5}}}&{\underline{\sf{\:\:1125 \:\:}}}\\ {\underline{\sf{5}}}& \underline{\sf{\:\:225 \:\:}} \\\underline{\sf{3}}&\underline{\sf{\:\:45\:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:15 \:\:}} \\ {\underline{\sf{5}}}& \underline{\sf{\:\:5\:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

$\red{\rm :\longmapsto\:Prime \: factorization \: of \: 1125 = {5}^{3} \times {3}^{2} }$

Now, we have

$\red{\rm :\longmapsto\:Prime \: factorization \: of \: 1125 = {5}^{3} \times {3}^{2} }$

$\red{\rm :\longmapsto\:Prime \: factorization \: of \: 144 = {2}^{4} \times {3}^{2} }$

Hence,

$\bf :\longmapsto\:LCM \: ( \: 1125, \: 144 \: )$

$\rm \: \: = \: \: {2}^{4} \times {3}^{2} \times {5}^{3}$

$\rm \: \: = \: \: 16 \times 9 \times 125$

$\rm \: \: = \: \: 16 \times 9 \times 125$

$\rm \: \: = \: \: 18000$

Hence,

$\bf :\longmapsto\:LCM \: ( \: 1125, \: 144 \: ) = 18000$

Additional Information :-

Let us consider two positive integers x and y, then

• 1. HCF (x, y) × LCM (x, y) = x × y

• 2. HCF always divides x, y and LCM

• 3. HCF < or = (x, y)

• 4. LCM > or = (x, y).