Question

What is the smallest number which leaves remainder 3 ,when divided by any of the numbers 5,6 or 8 ,but leaves no remainder when it is divided by 9?

Answer 1

L.C.M. of 5, 6, 7, 8 = 840.

 Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

 Required number = (840 x 2 + 3) = 1683.

Answer 2

1. first find the lcm of the given 3 numbers 5,6 and 8
2. the lcm is 120, if we add 3 to 120 it becomes 123, but 123 is not divisible by 9.
3. now we take the next multiple of 120 that is 240, add 3 and check divisibility by 9
4. it is not divisible
5. we repeat this process till we find a suitable number
6. the answer to this question is 843.