Find the largest possible $4$ -digit number divisible by $45$, $18$ and $35$.

A number that is divisible by $45$, $18$, and $35$ must also be divisible by their Lowest Common Multiple ($\text{LCM}$).
Let's find the $\text{LCM}$ of $45$, $18$, and $35$:

$$45=3^2\times 5$$

$$18=2\times 3^2$$

$$35=5\times 7$$

$$\text{LCM}=2\times 3^2\times 5\times 7=2\times 9\times 5\times 7=630$$

The largest possible $4$ -digit number is $9999$. We divide this number by $630$ to find the remainder:

$$9999÷630=15\text{ with a remainder of }549$$

Subtracting the remainder from $9999$ gives the largest $4$ -digit multiple of $630$:

$$\text{Required Number}=9999-549=9450$$