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
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
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.
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