Determine the smallest $3-$ digit number which is exactly divisible by $6,8$ and $12$

Smallest number divisible by $6,8,12=$L.C.M of $6,8,12$

Factors of $6=1\times 2\times 3$

Factors of $8=1\times 2\times 2\times 2$

Factors of $12=1\times 2\times 3\times 2$

So, L.C.M$=2\times 2\times 2\times 3=24$

$\therefore 24$ 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 $24$

$100=24\times 4+4$

So, remainder is $4$

But we need to find a number where remainder is $0$

$\therefore$ Required number$=100+(24-4)=100+20=120$

So,$120$ is the smallest $3$ digit number.