Question

Q. The smallest number, which when divided by 12,15,20 and 54 leaves in each case a reminder of 8 is:​

Answer 1

Answer:

Required number = (L.C.M. of 12, 15, 20, 54) + 8

= 540 + 8

= 548.

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L.C.M

2 | 12,15,20,54

2 | 6 ,15,10,27

3 | 3 ,15,5 ,27

5 | 1 ,5 ,5 ,9

1 ,1 ,1 ,9

NOW,2*2*3*5*9=540

Required number = (L.C.M. of 12, 15, 20, 54) + 8

= 540 + 8

= 548.

hope it's correct and helpful for you

Answer 2

Answer :-

Concept - If we have to find the smallest number which when divided by p,q & r leaves same remainder 'a' in each case, then

Required number = LCM ( p , q & r ) + 'a'

Solution :-

LCM ( 12 , 15 , 20 , 54 ) = 540

Required number = LCM ( 12 , 15 , 20 , 54 ) + 8

= 540 + 8

= 548

Hence, Required number = 548