smart but very slow solution...

This is wrong, the loop can go to n.
Test it with "print(i)" with n=7883 for example
7883 is a prime number, this code will test every number until 7883.

the loop actually never goes beyond sqrt(n)

'xrange' is a function in Python 2.x.
In Python 3.x, use range() function instead.

Excellent. Thank you for spelling this out for me.

NameError: name 'xrange' is not defined
'xrange' was renamed to 'range' in Python 3 appearently.

Also I would change n /= i to n //= i to more explicity call integer division, tho I'm not certain it matters since it's only running on modulo 0 stuff in the while loop, but normally it would cause some undesired non-integer results.

The given implementation can be improved by just changing the upperbound of the range to be int(math.sqrt(n))+2

Still probably not the most efficient approach, but there's no point in checking numbers above the sqrt(n)+1.

Edit:
Actually I think it does not matter since the return statement will exit the function before those higher values are tested.

This is a really smart solution, it takes advantage of the fact that the smallest factor of a number must be prime.

Your code is much more efiicient

I literally had the same piece of code but the fcking when I tried it, it would always timeout on me. Idk how it worked for you.

This may not work if the n is super big, the accuracy is limited by the digits of n.

A little improvization on this solution will take care of null factor. Check my solution.

What you said is correct, if the number has a large factor, it will be time consuming.Please refer to my solution.