Find the smallest number which when divided by $4$, $6$, $7$, $8$ and $9$ leaves the remainder $3$ in each case.

To find the smallest number which leaves a remainder of $3$ when divided by $4$, $6$, $7$, $8$, and $9$, we first need to find the Least Common Multiple ($\text{LCM}$) of these numbers and then add the remainder $3$ to it.
**Step 1: Find the $\text{LCM}$ of $4,6,7,8,9$ **

* Prime factorization of the numbers:

$$4=2^2$$

$$6=2\times 3$$

$$7=7^1$$

$$8=2^3$$

$$9=3^2$$

* Taking the highest power of each prime factor involved:

$$\text{LCM}(4,6,7,8,9)=2^3\times 3^2\times 7^1$$

$$\text{LCM}=8\times 9\times 7=72\times 7=504$$

Step 2: Add the required remainder

* Required Number $=\text{LCM}+\text{Remainder}$
* Required Number $=504+3=507$

Therefore, the smallest number is $507$.