Question

Find the least number which when divided by 7 ,14 , 21 leaving remainder 6 in each case. *​

Answer 1

Given :

7 ,14 , 21 leaving remainder 6 in each case.

To find :

Find the least number

Solution :

2 | 7 , 14 , 21

3 | 7 , 7 , 21

7 | 7 , 7 , 7

| 1 , 1 , 1

LCM = 2 x 7 x 3

LCM = 42

the required number = 42 + 6

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀ = 48

Extra information :

Product of two numbers = (their H.C.F.) × (their L.C.M.) .

LCM of fraction = LCM of numerator/HCF of denominator

HCF of fraction = HCF of numerator/LCM of denominator

LCM x HCF = products of the two numbers

Answer 2

7,14,21

Find put the LCM of these 3

$:\implies\sf 7= 1\times 7\\ \\ \\ :\implies\sf 14= 2\times 7\\ \\ \\ :\implies\sf 21= 3\times 7\\ \\ \\ :\implies\sf (LCM)= 1\times 2\times 3\times 7\\ \\ \\ :\implies\sf (LCM)= 42$

$\sf [required\ number]= (42+6)\\ \\ \\ :\implies\sf The \ number = 48$

$\underline{\boxed{\purple{\sf The \ least \ number = 48}}}$