Smallest number divisible by 6,8,12=L.C.M of 6,8,12
Factors of 6=1×2×3
Factors of 8=1×2×2×2
Factors of 12=1×2×3×2
So, L.C.M=2×2×2×3=24
∴24 is the smallest number divisible by 6,8,12
Now, we need to find the smallest 3− digit number divisible by 6,8,12
Smallest 3− digit number=100
We divide 100 by 24
100=24×4+4
So, remainder is 4
But we need to find a number where remainder is 0
∴ Required number=100+(24−4)=100+20=120
So,120 is the smallest 3 digit number.