## 5 Answers

$x \text{ and } y$ can be expressed as

- $x = 2^{a_{1}}\cdot 3^{a_{2}}\ldots P^{a_{n}}.$
- $y= 2^{b_{1}}\cdot 3^{b_{2}}\ldots P^{b_{n}}.$

where, $a_{i} \& b_{i} \geq 0$ for $1\leq i \leq n$, and $P$ is a prime number.

$x*y = 2^{a_{1}+b_{1}}\cdot 3^{a_{2}+b_{2}}\ldots P^{a_{n}+b_{n}}$

- $\text{LCM} (x,y) = 2^{\max(a_{1},b_{1})}\cdot 3^{\max(a_{2},b_{2})}\ldots P^{\max(a_{n},b_{n})}$
- $\text{HCF} (x,y) = 2^{\min(a_{1},b_{1})}\cdot 3^{\min(a_{2},b_{2})}\ldots P^{\min(a_{n},b_{n})}$

$\max(a_{i} ,b_{i}) + \min(a_{i},b_{i}) = a_{i} + b_{i}$

So, $\text{LCM}(x,y)\ast \text{HCF}(x,y) = 2^{a_{1}+b_{1}}\cdot 3^{a_{2}+b_{2}}\ldots P^{a_{n}+b_{n}} = x\ast y.$

Hence proved.

Let a ( 120 ) = f1 * f2 * f3 * f4 * f5 .... ( 2 * 2 * 2 * 3 * 5 ).

b ( 18 ) = F1 * F2 * F3........ ( 2 * 3 * 3 )

LCM of a( 120 ) and b( 18 ) = product of elements of a **union** b.

= M( {f1, f2, f3, f4, f5 } **U** { F1, F2, F3 } ) //M- > multiplication.

= M( (f1/F1), f2, f3, (f4/F2), F3, f5 ) ----- (A) // f1/ F1- > f1 or F1

HCF of a( 120 ) and b( 18 ) = product of elements of a **intersection** b.

= M( {f1, f2, f3, f4, f5 } **$\cap$** { F1, F2, F3 } )

= M( (f1/F1), (f4/F2) ) -----------------------(B)

// Remember if a **$\cap$ ****b****= ∅, **then HCF = 1.

From equation A and B,

Product of LCM and HCF = M( (**f1**/F1), f2, f3, (**f4**/F2), F3, f5 ) * M( (f1/**F1**), (f4/**F2**) )

= M( f1, f2, f3, f4, F3, f5, F1, F2 )

= Product of a and b.