Question

Find the LCM and HCF of integers and verify LCM × HCF = Product of two numbers

180 and 192

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$    

   

$\begin{array}{r | l} 2 & 180 \\ \cline{2-2} 2 & 90 \\ \cline{2-2} 3 & 45 \\ \cline{2-2} 3 & 15 \\ \cline{2-2} & 5 \end{array}$

$\begin{array}{r | l} 2 & 192 \\ \cline{2-2} 2 & 96 \\ \cline{2-2} 2 & 48 \\ \cline{2-2} 2 & 24 \\ \cline{2-2} 2 & 12 \\ \cline{2-2} 2 & 6 \\ \cline{2-2} & 3 \end{array}$

180 = 2² × 3² × 5

$\tt{\rightarrow 192=2^6\times 3^1}$

HCF(180,192) = 2² × 3¹ = 12

LCM(180,192)  

$\tt{\rightarrow 5\times 2^6\times 3^2}$

= 576 × 5  

= 2880

Verification

LCM × HCF = 2880 × 13 = 34560

180 × 192 = 34560

 

Hence Proved  

LCM × HCF = Product of two numbers

Answer 2

Answer:

${\boxed{\bold{L.C.M. =2880}}}$

${\boxed{\bold{H.C.F. =12 }}}$

Step-by-step explanation:

Given:-

• Numbers = 180 and 192
• L.C.M =?
• H.C.F. = ?

To verify:-

• LCM × HCF = Product of numbers

$\tt At\:first,$

$\rm 180 = 2 \times 90 \\\rm \: \: \: \: \: \: \:= 2 \times 2 \times 45 \\\rm \: \: \: \:\: \: \:= 2\times 2 \times 3 \times 15 \\\rm \: \: \: \:\: \: \: = 2 \times 2\times 3\times 3\times 5$

$\tt Secondly,$

$\rm 192 = 2\times 96 \\\rm \: \: \: \: \:\: \:= 2 \times 2\times 48 \\\rm \: \: \:\: \: \: \:= 2\times 2 \times 2\times 24 \\\rm \: \: \: \: \:\: \: = 2\times 2\times 2\times 2\times 12 \\\rm \: \: \: \: \: \:\:= 2 \times 2 \times 2\times 2\times 2\times 6 \\\rm \:\: \: \: \: \:\: = 2 \times 2\times 2\times 2 \times 2 \times 2\times 3$

$\rm LCM = {2}^{6} \times {3}^{2} \times 5 \\\rightarrow\rm LCM = 64 \times 9 \times 5 \\\rightarrow{\boxed{\rm {LCM = 2880}}}$

$\rm HCF ={2}^{2} \times 3 \\\rightarrow\rm HCF = 4 \times 3 \\\rightarrow{\boxed{\rm{HCF = 12}}}$

Verification:-

$\rm LCM \times HCF = Product\:of\:numbers \\\rightarrow\rm 2880 \times 12 = 180\times 192 \\\rightarrow{\boxed{\rm{34560 = 34560}}}$

$\therefore\bold{\underline{L.H.S = R.H.S \: \: \: [Verified]}}$