cool thanks thats what i changed it to than just now i changed to do something else and instead of leaving the program running for 4 or 5 days as i had been i got my answer in less than a second ^_^"
i realized that i was wasting way to much time checking every possible answer because the number is jumping around so much so instead i wrote a real quick program infact i wrote it in interactive that just starts at 1 and works the problem backwards for how ever long i tell it to :)) thanks for considering the homework, but this time you did'nt have to worry i did the homework last week i was just doing this for fun, it randomly asked something like "what about a number that takes 100 steps to get to 1" and i thought what about 1000 :)) now i have one running looking for a number that takes 1000000 steps unfortunately i dont think ill get any extra credit for doing any of this lol here is the program sorry copying it from interactive: >>> number=1 >>> count=1000 >>> count=0 >>> while count < 1000: ... if (number*2)%2 == 0: ... number=number*2 ... else: ... number=(number-1)/3.0 ... count=count+1 ... print number On Sep 4, 2007, at 3:00 AM, Alan Gauld wrote: > Max, > >> just wondering if theirs a way to check if a larger number is even >> or >> odd > > I notice this is homework so I won't give a direct answer. > But look at the mod operator '%' - It returns the remainder > of an integer division. An even number is exactly divisible > by 2. > > HTH, > > Alan G. > > "max baseman" <[EMAIL PROTECTED]> wrote > >> about before ive let it sit for a few days now and i reached a >> number >> to high to convert to a decimal by adding 0.0 > > Thats a terrible way to convert to a float. Just use the float > function > (actually a type): > > f = float(anInt) > > But to determine if its odd/even you don't need to use floats at all. > > >> here's the program: >> >> count=1 >> numstart=268549802 >> number=0 >> high=0 >> a=0 >> while 1==1: >> numstart=numstart+1 >> number=numstart >> count=1 >> while number !=1: >> if number/2 == (number+0.0)/2: >> number=number/2 >> else: >> number=(number*3)+1 >> count=count+1 >> if count > a: >> a=count >> print numstart,":",count > > > Hmm, I have no idea what this is doing... > >> >> thanks >> >> _______________________________________________ >> Tutor maillist - Tutor@python.org >> http://mail.python.org/mailman/listinfo/tutor >> > > > _______________________________________________ > Tutor maillist - Tutor@python.org > http://mail.python.org/mailman/listinfo/tutor _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor