To multiply by any value of 2 to the power of N (i.e. 2^N) shift the bits N times to the left.
0000 0001 = 1 times 4 = (2^2 => N = 2) = 2 bit shift : 0000 0100 = 4 times 8 = (2^3 -> N = 3) = 3 bit shift : 0010 0000 = 32
etc..
To divide shift the bits to the right.
The bits are whole 1 or 0 – you can’t shift by a part of a bit thus if the number you’re multiplying by is does not factor a whole value of N ie.
since: 17 = 16 + 1 thus: 17 = 2^4 + 1 therefore: x * 17 = (x * 16) + x in other words 17 x's
thus to multiply by 17 you have to do a 4 bit shift to the left, and then add the original number again:
==> x * 17 = (x * 16) + x ==> x * 17 = (x * 2^4) + x ==> x * 17 = (x shifted to left by 4 bits) + x so let x = 3 = 0000 0011 times 16 = (2^4 => N = 4) = 4 bit shift : 0011 0000 = 48 plus the x (0000 0011)
ie.
0011 0000 (48)
+ 0000 0011 (3)
=============
0011 0011 (51)
Edit: Update to the original answer. Charles Petzold has written a fantastic book ‘Code’ that will explain all of this and more in the easiest of ways. I thoroughly recommend this.