The greatest number of four digits which is divisible by 18, 25, 45 and 55 is:

Correct option is A

Given:
We need to find the greatest four-digit number divisible by 18, 25, 45, and 55.
Solution:
Prime factorization:
18 =$$2 \times 3^2$$​
25 = $$5^2$$​​
45 $$= 3^2 \times 5$$​
55 =$$5 \times 11$$​
LCM = $$2 \times 3^2 \times 5^2 \times 11 = 2 \times 9 \times 25 \times 11 = 4950$$

​​$$\frac{9999}{4950} \approx 2.02$$​​
The largest integer less than or equal to 2.02 is 2.
Multiply 4950 by 2:
$$4950 \times 2 = 9900$$​​
The greatest four-digit number divisible by 18, 25, 45, and 55 is 9900.