Divide arbitrary binary integers by 3 in python

Question

All the binary integers am given are always divisible by 3 so no floating-point values involved. The return value must also be a binary string Tried this but a and h are different yet they should be

>>> a="1111111111111111111111111111111111111111111111111111111111111111111111"
>>> f=int(a,base=2)
>>> print (f)
1180591620717411303423
>>> g = f/3
>>> c=int(g*3)
>>> print(c)
1180591620717411303424
>>> h = bin(c)
>>> print(h)
0b10000000000000000000000000000000000000000000000000000000000000000000000

Answer 1

Since your numbers are always divisible by 3 it is better to use // for integer division, because when you use floating point division for large numbers the result might not be exactly right.

So your code will be something like:

>>> a="1111111111111111111111111111111111111111111111111111111111111111111111"
>>> f=int(a, base=2)
>>> print(f)
1180591620717411303423
>>> g = f // 3
>>> c = g * 3
>>> print(c)
1180591620717411303423
>>> h = bin(c)
>>> print(h)
0b1111111111111111111111111111111111111111111111111111111111111111111111