What is the smallest number which when decreased by $8$ is exactly divisible by $21$, $27$, $33$ and $55$?

The required number is $\text{LCM}(21,27,33,55)+8$.
Let us find the prime factorization of each number:

* $21=3\times 7$
* $27=3^3$
* $33=3\times 11$
* $55=5\times 11$

Now, take the highest power of each prime factor present:

$$\text{LCM}=3^3\times 5\times 7\times 11$$

$$\text{LCM}=27\times 5\times 7\times 11$$

$$\text{LCM}=135\times 77=10395$$

Add $8$ to the LCM to find the final number:

$$\text{Number}=10395+8=10403$$

Since $10403$ is not given in choices A, B, or C, the correct option is "None of the above".