What is the least number which when divided by $12$, $20$ and $24$ leaves in each case a remainder of $8$?

1. To find the required number, we first need to find the Least Common Multiple ($\text{LCM}$) of $12$, $20$, and $24$.
* Prime factorization of $12=2^2\times 3$
* Prime factorization of $20=2^2\times 5$
* Prime factorization of $24=2^3\times 3$

2. Calculate the $\text{LCM}$:

$$\text{LCM}(12,20,24)=2^3\times 3\times 5=8\times 3\times 5=120$$

3. Since the division leaves a remainder of $8$ in each case, add the remainder to the $\text{LCM}$:

$$\text{Required Number}=\text{LCM}+\text{Remainder}=120+8=128$$