Too big Number, isn't it?

You can solve this problem if you know this rule about remainders. Let a number $x$ divide the product of A and B. The remainder will be the product of the remainders when $x$ divides A and when $x$ divides B.

Using this rule,

The remainder when 33 divides 1044 is 21.

The remainder when 33 divides 1047 is 24.

The remainder when 33 divides 1050 is 27.

The remainder when 33 divides 1053 is 30.

$∴$ The remainder when 33 divides $1044 \times 1047 \times 1050 \times 1053$ is $21 \times 24 \times 27 \times 30$ .

Note : The remainder when a number is divided by a divisor $d$ will take values from 0 to $(d-1)$ . It cannot be equal to or more than $d$ .

The value of $21 \times 24 \times 27 \times 30$ is more than $33$ . When the value of the remainder is more than the divisor, final remainder will be the remainder of dividing the product by the divisor.

That is, the final remainder is the remainder when 33 divides $21 \times 24 \times 27 \times 30$ . When 33 divides $21 \times 24 \times 27 \times 30$ , the remainder is 30.

Thus, the remainder is $\boxed{30}$

$1044\cdot 1047\cdot 1050\cdot 1053\equiv 21\cdot 24\cdot 27\cdot 30\equiv 30\quad \left( \text{mod 33} \right)$

Note: $21\cdot 24\cdot 27\cdot 30$ can be grouped as $\left( 21\cdot 24 \right) \left( 27\cdot 30 \right) =504\cdot 810$ to simplify the answer.