Question

logether?
Find the smallest 4 digit numbers
such that when it is divided by 12
18, 21 and 28, it leaves remainder
3 in each case.​

Answer 1

The samllest 4 digit number is 1008

Answer 2

Answer:

$\blue\dag$ The required number is 1011.

SOLUTION

To find the smallest  4 digit numbers such that when it is divided by 12 ,18, 21 and 28 and leaves remainder 3 in each case, Let us find the LCM of 12,18,21,28.

LCM

To find the LCM we must do the prime factorization of the given numbers.

12 = 2 × 2 × 3

18 = 2 × 3 × 3

21 = 3 × 7

28 = 2 × 2 × 7

∴ LCM of 12 ,18, 21 and 28  = 2 × 2 × 3 × 3 × 7

⇒ LCM of 12 ,18, 21 and 28  = 252.

∵ It is a 4 digit number the required number will be 252 × 4 = 1008(Smallest 4 - digit number that is divisible by 252)

∵ It is also said in the question that  it leaves remainder 3 in each case.​

The required number = 1008 + 3

⇒ The required number = 1011