Math, asked by rkpandey14937442, 1 year ago

find the smallest 4 digit number such that when it is divided by 12, 18, 21 and 28, it leaves remainder 3 in each case​

Answers

Answered by shivi001
82

take the LCM of 12 ,18,21,28 we get 252

and smallest 4 digit no is 1000 so

252×4=1008 but reminder should be 3 so we add 3 in this no

required answer is 1011

Answered by JeanaShupp
44

The required number is 1011.

Explanation:

Since  , 12 = 2 x 2 x 3

18 = 2 x 3 x 3

21 = 3 x 7

28 = 2 x 2 x 7

Least common multiple of 12, 18, 21 and 28 =   2 x 2 x 3 x 3x 7 =252

252 x 2 = 504

252 x 3 =756

252 x 4= 1008

Thus, the smallest 4 digit number such that when it is divided by 12, 18, 21 and 28 is 1008.

Also, the required no. leaves remainder 3.

So , the required number = 1008+3 = 1011

Hence, the required number is 1011.

# Learn more :

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