On Sun, Jan 17, 2016 at 9:23 AM, <jonas.thornv...@gmail.com> wrote: > function factor_it(i){ > prime=true; > sqroot=Math.floor(Math.sqrt(i)); > for (j=2;j<sqroot;j++){ k=i/j; prime=!Number.isInteger(k); if (prime) {return > prime}} > return prime; > }
A couple of potential problems here. The first thing that comes to mind is that floating point inaccuracy is going to bite you long before the numbers "seem huge" to someone who's thinking about 2**53. The second is an off-by-one error: a perfect square may come up as prime. Check for those and see how it looks. Also, check your double-use of the 'prime' variable, which also appears to be global here. ChrisA -- https://mail.python.org/mailman/listinfo/python-list