#!/usr/bin/python
from primes_sieve import primes
from tools import isPandigital
# Tried with 987654321 but was too big.
# Went down until 7654321 and got the answer.
primes = primes(7652413)
for prime in primes[::-1]:
if isPandigital(prime, 7):
print prime
break
#! /usr/bin/python
from primes_sieve import primes, isPrime
def truncate(n, direction):
if len(str(n)) == 1 and isPrime(n):
return True
elif isPrime(n):
if direction == 'LR':
num = int(str(n)[1:])
elif direction == 'RL':
num = int(str(n)[:-1])
return truncate(num, direction)
return False
# Generate a primes list. This is a bad
# idea and could be optimized much more.
p = primes(800000)
# The [4:] removes the unwanted values
# of 2, 3, 5 and 7.
result = [i for i in p if truncate(i, 'LR') and truncate(i, 'RL')][4:]
print result
print sum(result)
#! /usr/bin/python
from primes_sieve import primes
primes = primes(9999)
# The next prime after 1487
primes = primes[primes.index(1493):]
for a in primes:
b, c = a+3330, a+6660
if b in primes and c in primes:
a, b, c = str(a), str(b), str(c)
if sorted(a) == sorted(b) == sorted(c):
print '{0}{1}{2}'.format(a,b,c)
break
#!/usr/bin/python
from primes_sieve import primes, isPrime
from time import time
start = time()
limit = 10**6
p = primes(limit)
prev_longest = 0
for i in xrange(0,len(p)):
s, counter = 0, 0
for num in p[i:]:
s += num
counter += 1
if isPrime(s) and counter > prev_longest and s <= limit:
prev_longest = counter
print s, '| Terms:',counter
elif s > limit: break
end = time()
print end-start
#!/usr/bin/python
from primes_sieve import primes
#####
# This problem needs a facelift.
# setting the bounds for the exponent and
# for the primes list is not efficient.
# A more efficient solution would implement a while
# loop until the smallest number is encountered.
#####
primes, exp = primes(6000), 40
goldbach = set(p + 2 * (i**2) for p in primes for i in xrange(1, exp))
nonprimes = [n for n in xrange(1, primes[-1]) if n not in primes]
result = set()
for n in nonprimes:
for j in xrange(1, exp):
x = n + 2 * (j**2)
if x not in goldbach and x not in primes and x % 2 != 0:
result.add(x)
print min(result)
#! /usr/bin/env python from primes_sieve import primes #The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. #Find the sum of all the primes below two million. c = 0 for i in primes(2000000): c += i print c
#! /usr/bin/env python
from primes_sieve import primes
#The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
#Find the sum of all the primes below two million.
c = 0
for i in primes(2000000):
c += i
print c
#! /usr/bin/env python from primes_sieve import primes print primes(105000)[10000]
#!/usr/bin/python
from primes_sieve import primes
#####
# This problem needs a facelift.
# setting the bounds for the exponent and
# for the primes list is not efficient.
# A more efficient solution would implement a while
# loop until the smallest number is encountered.
#####
primes, exp = primes(6000), 40
goldbach = set(p + 2 * (i ** 2) for p in primes for i in xrange(1, exp))
nonprimes = [n for n in xrange(1, primes[-1]) if n not in primes]
result = set()
for n in nonprimes:
for j in xrange(1, exp):
x = n + 2 * (j ** 2)
if x not in goldbach and x not in primes and x % 2 != 0:
result.add(x)
print min(result)
#! /usr/bin/python from primes_sieve import primes p = primes(100) pointer = 0 d = [] for i in p: if sum(p[:pointer]) in p: d.append(sum(p[:pointer])) print p[:pointer],sum(p[:pointer]) pointer += 1 print d
#! /usr/bin/python
from primes_sieve import primes, isPrime
def truncate(n, direction):
if len(str(n)) == 1 and isPrime(n):
return True
elif isPrime(n):
if direction == 'LR':
num = int(str(n)[1:])
elif direction == 'RL':
num = int(str(n)[:-1])
return truncate(num, direction)
return False
# Generate a primes list. This is a bad
# idea and could be optimized much more.
p = primes(800000)
# The [4:] removes the unwanted values
# of 2, 3, 5 and 7.
result = [i for i in p if truncate(i, 'LR') and truncate(i, 'RL')][4:]
print result
print sum(result)