def problem49():
"""
The arithmetic sequence, 1487, 4817, 8147, in which each of the terms
increases by 3330, is unusual in two ways: (i) each of the three terms
are prime, and, (ii) each of the 4-digit numbers are permutations of
one another.
There are no arithmetic sequences made up of three 1-, 2-, or 3-digit
primes, exhibiting this property, but there is one other 4-digit
increasing sequence.
What 12-digit number do you form by concatenating the three terms in
this sequence?
"""
limit = 10000
primes, primeset = util.primes(limit)
for a in primes:
permset = set([int(x) for x in util.permutations(str(a))])
for b in util.permutations(str(a)):
b = int(b)
if b > a and b in primeset:
c = b + (b-a)
if c in primeset and c in permset and c != 8147:
return "{}{}{}".format(a, b, c)
def problem43():
"""
The number, 1406357289, is a 0 to 9 pandigital number because it is made
up of each of the digits 0 to 9 in some order, but it also has a rather
interesting sub-string divisibility property.
d2d3d4=406 is divisible by 2
d3d4d5=063 is divisible by 3
d4d5d6=635 is divisible by 5
d5d6d7=357 is divisible by 7
d6d7d8=572 is divisible by 11
d7d8d9=728 is divisible by 13
d8d9d10=289 is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.
"""
factors = 2, 3, 5, 7, 11, 13, 17
total = 0
for digits in util.permutations('0123456789'):
for i in xrange(1, 8):
if int(digits[i:i+3]) % factors[i-1] != 0:
break
else:
total += int(digits)
return total
def problem41():
"""
We shall say that an n-digit number is pandigital if it makes use of
all the digits 1 to n exactly once. For example, 2143 is a 4-digit
pandigital and is also prime.
What is the largest n-digit pandigital prime that exists?
"""
from random import randint
def is_prime(n):
if pow(2, n, n) != 2:
return False
for i in xrange(50):
a = randint(2, n-1)
if util.gcd(n, a) == 1 and pow(a, n-1, n) != 1:
return False
return True
best = 2143
for j in xrange(1, 10):
for i in util.permutations(''.join(map(str, xrange(1, j+1)))):
i = int(i)
if i > best:
if is_prime(i):
best = i
return best
def problem24():
"""
What is the millionth lexicographic permutation of the digits
0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
"""
limit = 1000000
permute = util.permutations('0123456789')
for j in xrange(limit - 1):
next(permute)
return next(permute)
def problem32():
"""
We shall say that an n-digit number is pandigital if it makes use of all
the digits 1 to n exactly once; for example, the 5-digit number, 15234,
is 1 through 5 pandigital.
The product 7254 is unusual, as the identity, 39 x 186 = 7254, containing
multiplicand, multiplier, and product is 1 through 9 pandigital.
Find the sum of all products whose multiplicand/multiplier/product
identity can be written as a 1 through 9 pandigital.
"""
product_set = set()
for p in util.permutations('123456789'):
for i in xrange(1, len(p) / 3 + 1):
for j in xrange(i + 1, len(p) / 3 * 2 + 1):
triplet = int(p[:i]), int(p[i:j]), int(p[j:])
if triplet[0] * triplet[1] == triplet[2]:
product_set.add(triplet[2])
return sum(product_set)
def prime_perms(r):
for perm in permutations(range(1, r)):
n = int(''.join(str(c) for c in perm))
if is_prime(n):
yield n
if len(sys.argv) != 2:
usage()
try:
n = int(sys.argv[1])
except ValueError:
usage()
l = len(tokens)
limit = 2**18
if n < 0 or n > l:
usage()
if factorial(l) / factorial(l-n) <= limit:
for s in util.permutations(n, tokens):
print(''.join(s))
else:
strings = set()
while len(strings) < limit//2:
new = ''
choices = list(tokens)
for i in range(n):
token = random.choice(choices)
choices.remove(token)
new += token
if tokens[0] in new:
strings.add(new)
