At the very least, you could do it with logs like so: >>> from math import exp, log >>> ref = [i for i in range(1, 500000)] >>> test = [i if i !=

At the very least, you could do it with logs like so: >>> from math import exp, log >>> ref = [i for i in range(1, 500000)] >>> test = [i if i != 7922 else 59148 for i in ref] >>> s = sum(test) - sum(ref) >>> r = exp(sum(map(log, test)) - sum(map(log, ref))) >>> b = s/(r - 1); b 7921.999994157095 >>> (int(round(b)), s + int(round(b))) (7922, 59148) The use of logs is really just a convenience for finding product(test) and product(ref), so that we have both a - b (from the sums) and a/b (from the ratio of the products). Python of course has bignums, so we can also calculate these products directly, but that gets to be a very large calculation; you don't want to be storing N! for N ~= 2^64 if at all possible. If we know that these numbers are bounded by 2^n (e.g. they're all longs) then we might be able to do this product modulo some larger base, perhaps even 2^n, with the assurance that when we do the division all of that crap will cancel; I don't know and haven't tried it.