Thread: Division in binary (repeated subtraction method)

Divide 1000000001 (base 2) by 10011 (base 2)

answer = 11011 using repeated subtraction but I have no idea what is going on, here is a link to the pdf exam paper and look at the answer to question 1) (b) iii) LINK - can somebody please explain what is going on here???

First Round is easy

0111111111
0000010011

Just subtract bit-wise

0111101100 - That's 01

It's the second round where it gets interesting.

0111101100
0000010011

The problem starts right off. You must borrow from two digits left, giving

011110101(10)
0000010011

There's another problem later on. Borrow again.

01110(10)0101(10)
0000010011

Now we're ready to subtract.

0111011001 - That's 10

It's tedious, but it's not tricky for any other reason.