On Sat, Jan 3, 2009 at 9:43 PM, Benjamin Serrato <benjamin.serr...@gmail.com> wrote: > I changed it to this: > > #!/usr/local/bin/python3.0 > #problem1.py > > """Lists the sum of all multiples of 3 and 5 less than 1000.""" > > def sumTotal(multipleInitial, multiple): # This awkward solution because > total = 0 # > mulitipleInitial = multiple just points > n = 2 # a new > variable to the same memory
You don't have to pass the duplicate arguments. It's true that multipleInitial=multiple makes multipleInitial reference the same value as multiple, but multiple = multipleInitial * n make mulltiple refer to a new value while leaving multipleInitial alone. This might help: http://personalpages.tds.net/~kent37/kk/00012.html > while multiple < 1000: # space > total = total + multiple > multiple = multipleInitial * n > n = n + 1 > return total > > print(sumTotal (3, 3) + sumTotal(5,5)) > > ----- > I think it does what I wanted it to do, but Kent pointed out I wanted > it to do was a false solution. So, I can create a list of all common > multiples below 1000, sum them, subtract them from 266 333. Or, what > I prefer, create a list of all multiples, checking against that list > for a multiple before adding a new multiple. But, I don't know how to > work with lists so I'll be back in a day or two. You might want to look at the sum() and range() functions. Kent PS Please subscribe to the list. _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
