There is only a simple way to find modulo of 2^i numbers using bitwise.
There is an ingenious way to solve Mersenne cases as per the link such as n % 3, n % 7... There are special cases for n % 5, n % 255, and composite cases such as n % 6.
For cases 2^i, ( 2, 4, 8, 16 ...)
n % 2^i = n & (2^i - 1)
More complicated ones are hard to explain. Read up only if you are very curious.