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

To find the largest 4-digit number divisible by 45, 18, and 35, we first determine their LCM:
$$45=3^2\times 5$$

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

$$35=5\times 7$$

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

The largest possible 4-digit number is $9999$. Dividing $9999$ by $630$:

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

To get the largest completely divisible 4-digit number, subtract the remainder from $9999$:

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

Alternatively, $630\times 15=9450$.