You can decide whether a number is divisible by 3 by forming its cross sum,
that is the sum of all its decimal digits: It is divisible by three exactly if its cross sum is. Similarly, a three-digit number is
divisible by seven if (and only if) the sum of twice its most significant
digit plus three times its middle digit plus its least significant digit is
divisible by seven.

Can you find a similar scheme for checking divisibility by 33 and 37, for numbers with an arbitrary number of digits?

Check if it's divisible by 3 (sum its digits and see if the sum is a multiple of 3) and by 11 (sum digits in odd positions, subtract digits in even positions, see if the result is a multiple of 11).

EXAMPLE: 1089: 1+0+8+9=18, multiple of 3. 1-0+8-9=0, multiple of 11. So, 1089 is multiple of 33.