Solution to problem 43.
"""
from utils.misc import build_int, prev_perm
def substring_divisible(lst):
if not build_int(lst[1:4]) % 2 == 0:
return False
if not build_int(lst[2:5]) % 3 == 0:
return False
if not build_int(lst[3:6]) % 5 == 0:
return False
if not build_int(lst[4:7]) % 7 == 0:
return False
if not build_int(lst[5:8]) % 11 == 0:
return False
if not build_int(lst[6:9]) % 13 == 0:
return False
if not build_int(lst[7:10]) % 17 == 0:
return False
return True
sum = 0
perm = range(9, -1, -1)
while perm[0] != 0:
if substring_divisible(perm):
sum += build_int(perm)
perm = prev_perm(perm)
print sum
"""
from utils.misc import build_int, prev_perm
def substring_divisible(lst):
if not build_int(lst[1:4]) % 2 == 0:
return False
if not build_int(lst[2:5]) % 3 == 0:
return False
if not build_int(lst[3:6]) % 5 == 0:
return False
if not build_int(lst[4:7]) % 7 == 0:
return False
if not build_int(lst[5:8]) % 11 == 0:
return False
if not build_int(lst[6:9]) % 13 == 0:
return False
if not build_int(lst[7:10]) % 17 == 0:
return False
return True
sum = 0
perm = range(9, -1, -1)
while perm[0] != 0:
if substring_divisible(perm):
sum += build_int(perm)
perm = prev_perm(perm)
print sum
# Author: Deddryk
"""
Solution to problem 41.
This is solved by checking every permutation of the digits 1..n
for n in 9..2 for primality.
"""
from utils.primes import probable_prime
from utils.misc import prev_perm, build_int
x = range(9, 0, -1)
for num_digits in range(9, 1, -1):
x = range(num_digits, 0, -1)
while x != range(1, num_digits + 1):
if probable_prime(build_int(x)):
print(build_int(x))
exit()
x = prev_perm(x)
# Author: Deddryk
"""
Solution to problem 41.
This is solved by checking every permutation of the digits 1..n
for n in 9..2 for primality.
"""
from utils.primes import probable_prime
from utils.misc import prev_perm, build_int
x = range(9, 0, -1)
for num_digits in range(9, 1, -1):
x = range(num_digits, 0, -1)
while x != range(1, num_digits + 1):
if probable_prime(build_int(x)):
print(build_int(x))
exit()
x = prev_perm(x)