HCF and LCM

HCF = 2 x 17 = 34

 The HCF of two consecutive even prime numbers is 2. 

let the number be x and 4x84*21=x*4x4(x^2)=1764x^2 =441x=214x=4*21=84

product of number =lcm*hcf480*x=2600*16 x=80

let no be 7x and 8xhence hcf=xlcm*hcf=product of two numbers280*x=56(x^2)x=5

Prime factorisation of 44 = 2 ^_$\times$ 2 ^_$\times$ 11 = 2^{2 _$\times$} 11¹ Prime factorisation of 90 = 2 ^_$\times$ 3 ^_$\times$ 3 ^_$\times$ 5 = 2^{1 _$\times$} 3^{2 _$\times$} 5¹

$\therefore$  LCM (44, 90) = 2^{2 _$\times$} 3^{2 _$\times$} 5^{1 _$\times$} 11¹ = 4 _^$\times$ 9 _^$\times$ 5 _^$\times$ 11 = 1980

The prime factorisation of 410 = 2^{1 _$\times$} 5^{1 _$\times$} 41 The prime factorisation of 96 = 2^{5 _$\times$} 3

  ||| |---|---| | 2| 96| | 2| 48| | 2| 24| | 2| 12| | 2| 6| | 3| 3| | | 1|

HCF = Product of the smallest powers of each common prime factor in the numbers HCF (410, 96) = 2¹ = 2

The prime factorisation of 153 = 3 $\times$ 3 $\times$ 1 ||| |---|---| | 3| 153| | 3| 51| | 17| 17| | | 1|   ||| |---|---| | 17| 4913| | 17| 289| | 17| 17| | | 1|

The prime factorisation of 4913 = 17 $\times$ 17 $\times$ 17 H.C.F (153, 4913) = product of the smallest powers of each common prime factor in the numbers.             = 17

Let x be the second number. H.C.F. ^_$\times$ L.C.M. = Product of two numbers 12 ^_$\times$ 240 = 60 ^_$\times$ x $\Rightarrow$ x = $\frac{240 \times 12}{60}$ = 48

Taking out the factors of the given numbers,

$36 = 2 \times 2 \times 3 \times 3$
$72 = 2 \times 2 \times 2 \times 3 \times 3$
$\therefore $ L.C.M of $36$  and $72$  $= 2 \times 2 \times 2 \times 3 \times 3 = 72$

Factors of the given numbers are,

Factors are $ 17 = 1 \times 17$
$5 = 1 \times 5 $
$\therefore $ LCM of $17$ and $5$  $= 1 \times 17 \times 5 = 85$

LCM of two co -prime numbers is their product.
Example: Consider $6$ and $7,$
Multiple of $6$ $ = 6,12,18,24,30,36,42, 48$
Multiple of $7$ $= 7,14,21,28,35,42$
L.C.M of $6$ and $7$ $= 42$
The product of $6$ and $7$ $= 6 \times 7 = 42$

$LCM \times HCF =$ product of numbers