On 13/08/13 17:45, #PATHANGI JANARDHANAN JATINSHRAVAN# wrote:
def check(num):
lis=[20,19,18,17,16,14,13,11] #because a no. divisible by 20 is
for i in lis:
if num%i==0:
continue
else:
return False
break
return True
This is a bit convoluted. All you need is
for i in lis:
if num%i != 0:
return False
return True
def main():
num=2520 # Because we can start from the no. divisible by 1-10
while not check(num):
print(num)
num+=2 # Because num has to be even
Surely you can increment by 20 since that's the next number that is
divisible by 20.
eg if we start at 20 the next number divisible by 20 will be 40, not
22... That should speed up the search considerably!
But my brain is nagging me that there should be a
cleverer mathematical trick too, but I can't remember
what it is... I'm sure one of our math gurus will tell us! :-)
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
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