Euler 46 Python

gistfile1.txt

# set up prime seive
n=10000
myPrimes = []
myPrimes = [True]*10001
myPrimes[0] = False
myPrimes[1] = False
for i in range (2,n):
	if myPrimes[i] == True:
		j = 2*i
		while j<=n:
			myPrimes[j]=False
			j=j+i
			
# set up set of squares
s = set()
for k in range (0,100):
	s.add(2*k*k)
	
# start searching up- test all odd non-prime numbers
i = 33
found = False
while found == False:
	if myPrimes[i] == False:
		for l in range (0,len(myPrimes)-1):
			if myPrimes[l]==True: 
				found = True
				test = i - l 
				if test in s:
					found = False
					break
		if found == True:
			print "Answer:", i
	i = i+2
		
	

i figured that out without asking djacobs! Jake Dobkin Gothamist.com (p) 917 627 6915 (f) 646 349 3893 AIM, YIM, GTalk: jakedobkin Mediakit: www.gothamistllc.com

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On Fri, Dec 9, 2011 at 5:19 PM, nataliepo < ***@***.*** > wrote: i am impressed with the array allocation syntax on line 4, jake! --- Reply to this email directly or view it on GitHub: https://gist.github.com/1453136