Question

hey guys
i know i'm gonna look very stupid but can anybody tell me what's wrong with this code:
can = 5
loop = 1
c = 1
while loop <5:
for m in range (2,can):
if can % m == 0 :
c = int (1)
break
c = str(1)
if c == str(1) :
print can
can = can +2
loop = loop +1
else:print ' not candidate' , can
can = can + 2
print 'the tenth prime is ' , can

Answer 1

Don't ever worry about looking stupid. We are all at different skill levels. As for your question I have no idea because I just started this class, so I am even stupider :)

Answer 2

you're right (: no one's stupid and i hope that i solve this

Answer 3

what are you trying to do?

http://pastebin.com/UtQG1PHp
using a code pasting site to post your code helps:
pastebin.com, dpaste.com, codepad.org, ideone.com - i bet there are a zillion of them

Answer 4

this code is wrong but what i'm trying to do is to compute and print the 1000'th prime number

Answer 5

well lines 5- 8 are a good start, they will figure out if can is prime of not and you preserve that information with c so you can use it later. the problem must be later.
an easier way to use c might be to give it a value of True or False - something like this
http://pastebin.com/FPhfd4qb

Answer 6

yep, lines 5- 8 work
http://pastebin.com/hQUBiriN

Answer 7

ok i think i've got it but it keeps adding 2 on the last prime:
can = 3
loop = 1
c = 1
while loop <10:
for m in range (2,can):
if can % m == 0 :
c = int (1)
break
c = str(1)
if c == str(1) :
print can
can = can +2
loop = loop +1
else:
print 'NOT CANDIDATE' , can
can = can + 2
print 'the tenth prime is ' , can

Answer 8

this is suppost to compute the 10"th prime

Answer 9

I wonder why it keeps adding two to the last prime?
you should use a code pasting site to post your code
http://pastebin.com/fVm5QXBq

Answer 10

thanks (:

Answer 11

Hi! It took quite long time, but... as I think I have figured out your problem: you missed '2' as prime, so you programm gives right answer for 10-th it is really 31(but if we exclude from sequence of prime numbers '2'). Desition of your problem: start loop from 2!

Answer 12

Also it is interesting to notice some fact, while cheking if number is prime:
a=b*c+r a is not prime if it can be writen as b*c (both integer) and r-reminder is zero

Answer 13

so as we know b>2 and c>2 due to them both integer and we check only odd numbers for prime, as even (all except 2) were eliminated as they have 2 as divider.
So form b>2 and c>2 we get that c=a/b where b>2, so instead checking range from 2 to prime-1 we can check range from 2 to prime/2. Result will be the same/ Think about it ))

Answer 14

http://pastebin.com/MrnXwkJS