Find the smallest number which will be completely divisible by $24,48$ and $60$.

To find the smallest positive non-zero number completely divisible by $24,48,$ and $60$, we need to find their Least Common Multiple (LCM).
Let's find the prime factorization of each number:

$$24=2^3\times 3^1$$

$$48=2^4\times 3^1$$

$$60=2^2\times 3^1\times 5^1$$

To find the LCM, we take the highest power of each prime factor present in these factorizations:

* Highest power of $2$ is $2^4$
* Highest power of $3$ is $3^1$
* Highest power of $5$ is $5^1$

$$\text{LCM}=2^4\times 3^1\times 5^1$$

$$\text{LCM}=16\times 3\times 5$$

$$\text{LCM}=48\times 5=240$$

Therefore, the smallest positive number divisible by all three numbers is $240$.