def problem37():
found = set()
primes = iPrime()
# The primes 2,3,5,7 are not candidates, but they could be the result of a
# truncation.
primesOfInterest = {next(primes), next(primes), next(primes), next(primes)}
for q in primes:
if isMadeOfOddOrWithLeading2(q):
primesOfInterest.add(q)
if all(p in primesOfInterest for p in leftTruncates(q)) and all(p in primesOfInterest for p in rightTruncates(q)):
found.add(q)
if len(found) > 10:
return sum(found)
def problem60():
primes = [3,7]
primeGen = iPrime()
next(primeGen); next(primeGen); next(primeGen); next(primeGen) # 2 and 5 are primes that will never concatinate
concatenate = {2:set(),3:set(),5:set(),7:{3}}
def concatenateWith(a):
concatenate[a] = {p for p in primes[:-1] if remarkable(a,p)}
return concatenate[a]
for a in primeGen:
primes.append(a)
A = concatenateWith(a)
for b in A:
B = concatenate[b] & A
for c in B:
C = concatenate[c] & B
for d in C:
D = concatenate[d] & C
for e in D:
return sum([a,b,c,d,e])
def problem7a(): ''' solution using an iterator ''' for index,p in enumerate(iPrime()): if index + 1 == 10001: return p
