Mod of power 2 on bitwise operators?

He meant that taking number mod 2^n is equivalent to stripping off all but the n lowest-order (right-most) bits of number.

For example, if n == 2,

number      number mod 400000001      0000000100000010      0000001000000011      0000001100000100      0000000000000101      0000000100000110      0000001000000111      0000001100001000      0000000000001001      00000001etc.

So in other words, number mod 4 is the same as number & 00000011 (where & means bitwise-and)

Note that this works exactly the same in base-10: number mod 10 gives you the last digit of the number in base-10, number mod 100 gives you the last two digits, etc.

What he means is that :

x modulo y = (x & (y − 1))

When y is a power of 2.

Example:

0110010110 (406) modulo0001000000 (64)  =0000010110 (22)^^^^<- ignore these bits

Using your example now :

1011000111011010 (45530) modulo0000000000000001 (2 power 0) =0000000000000000 (0)^^^^^^^^^^^^^^^^<- ignore these bits1011000111011010 (45530) modulo0000000000010000 (2 power 4) =0000000000001010 (10)^^^^^^^^^^^^<- ignore these bits

Consider when you take a number modulo 10. If you do that, you just get the last digit of the number.

  334 % 10 = 4  12345 % 10 = 5

Likewise if you take a number modulo 100, you just get the last two digits.

  334 % 100 = 34  12345 % 100 = 45

So you can get the modulo of a power of two by looking at its last digits in binary. That's the same as doing a bitwise and.