# The number 3797 has an interesting property. Being prime itself, it is
# possible to continuously remove digits from left to right, and remain prime
# at each stage: 3797, 797, 97, and 7. Similarly we can work from right to
# left: 3797, 379, 37, and 3.
#
# Find the sum of the only eleven primes that are both truncatable from left
# to right and right to left.
#
# NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
from p010 import prime_list3
from itertools import ifilter
primes = prime_list3(1000000)
prime_set = set(primes)
def truncations(number):
digits = str(number)
for i in xrange(1,len(digits)):
yield int(digits[:-i])
yield int(digits[i:])
def is_truncatable(number):
if number not in prime_set:
return False
if any( trunc not in prime_set for trunc in truncations(number) ):
return False
return True
# product of the coefficients, -79 and 1601, is -126479.
# Considering quadratics of the form:
# n^(2) + an + b, where |a| < 1000 and |b| < 1000
# where |n| is the modulus/absolute value of n
# e.g. |11| = 11 and |-4| = 4
# Find the product of the coefficients, a and b, for the quadratic expression
# that produces the maximum number of primes for consecutive values of n,
# starting with n = 0.
from p010 import prime_list3
primes = set(prime_list3(79 ** 2 + 999 * 79 + 1000))
def quadratic_fn(a, b):
def fn(n):
return n ** 2 + (a * n) + b
return fn
class Quadratic(object):
def __init__(self, a, b):
self.a = a
self.b = b
self.fn = quadratic_fn(a, b)
def pandigital_primes():
biggest_pandigital = 7654321
primes = prime_list3(biggest_pandigital+1)
for prime in ifilter(is_pandigital, primes):
yield prime
