Question

Determine the smallest 3 digit number which is exactly divisible by 6 8 and 12 and leaves remainder as 5

Answer 1

The answer must have all of the prime factors of 6, 8 and 12.

From 6, you get a 2 and a 3.

From 8 you get three 2's

From 12 you get two 2's and a three.

So to be divisible by each of 6, 8 and 12, a number must have one 3 and three 2's as factors, i.e. it must be divisible by 3*2^3 = 24.

So the number that we are looking for must be the smallest 3 digit number (i.e. greater than 99) divisible by 24.

That would be 5*24 = 120.