This is achieved using similar methods to those applied to long division of decimal numbers.
As with decimal division, we first select the minimum number of leading digits from the dividend, that will form a number greater than the divisor. We then divide this value by the divisor. This is actually simpler in binary, as the result will always be 1.
The smallest number we would divide by is 2, which in binary is 10. So in practice we would need to use 2 or more of the leading digits from the quotient, to form the first number that is larger than the divisor.
When this has been achieved, we write 1 in the quotient line above the lsb of the number that we formed from the leading bits of the dividend. We write the remainder of this step of the division below, with it’s lsb aligned with that of the number we formed. (The remainder is found by subtracting the divisor from the number we formed.)
We then carry down the next bit from the dividend, to append to the remainder, forming a new number.
If this number is smaller than the divisor (as it is in this case), we write a zero in the quotient line, above the digit we carried down.
We then repeat this step, until we have a number that is larger than the divisor.
When the number is larger than the divisor, we write a 1 in the quotient line, above the digit we carried down to form the number.
We then find the new remainder and write this below as before.
We continue this process until all the bits of the dividend have been included. (We have now reached this stage for the example above.)
Any remainder at this stage, is a final remainder from the division.
Therefore 11101/11 = 1001 remainder 10.
11101 bin = 29 decimal.
11 bin = 3 decimal.
29/3 = 9 r 2.
9 decimal = 1001 bin.
2 decimal = 10 bin.
Therefore 1100/10 = 110 remainder 0.
Check: 12/2 = 6 r 0, = 110 remainder 0