def all_prime(number):
l_to_r = range(1, len(number)-1)
r_to_l = range(2-len(number), 0)
for i in l_to_r:
if not is_prime(int(number[i:])):
return False
for j in r_to_l:
if not is_prime(int(number[:j])):
return False
return True
def main(n):
count = 0
i = 1
while count != n:
i += 1
if is_prime(i):
count += 1
return i
def main(n):
count = 0
i = 1
while count != n:
i += 1
if is_prime(i):
count += 1
return i
def main(n):
factors = []
n_sqrt = int(math.sqrt(n)) + 1
for i in range(2, n_sqrt):
if n % i == 0:
factors.append(i)
all_factors = [] + factors
for i in factors:
all_factors.append(n /i)
max_prime_factor = 0
for factor in all_factors:
if is_prime(factor) and factor > max_prime_factor:
max_prime_factor = factor
print max_prime_factor
def main(n):
factors = []
n_sqrt = int(math.sqrt(n)) + 1
for i in range(2, n_sqrt):
if n % i == 0:
factors.append(i)
all_factors = [] + factors
for i in factors:
all_factors.append(n / i)
max_prime_factor = 0
for factor in all_factors:
if is_prime(factor) and factor > max_prime_factor:
max_prime_factor = factor
print max_prime_factor
def all_prime(alters):
for alter in alters:
if not is_prime(int(''.join(alter))):
return False
return True
"""
from commons import is_prime
def all_prime(alters):
for alter in alters:
if not is_prime(int(''.join(alter))):
return False
return True
def circles(string):
"""make circle-numbers, which may be for general use later
e.g. 197's circle-numbers are 971, 719"""
numbers = []
current = string
for i in range(len(string)-1):
new = current[1:] + current[0]
numbers.append(new)
current = new
return numbers
result = []
for number in xrange(1000000):
if is_prime(number):
string = str(number)
alters = circles(string)
if all_prime(alters):
result.append(number)
print len(result)
print result
#!/usr/bin/env python """ Calculate the sum of all the primes below two million. """ from commons import is_prime print sum([i for i in xrange(2, 2000000) if is_prime(i)])
"""
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 itertools import permutations
from commons import is_prime
for n in range(9, 1, -1):
perms = permutations(range(n, 0, -1), n) # ensure firstly output larger ones
for p in perms:
str_p = [str(i) for i in p]
int_p = int(''.join(str_p))
if is_prime(int_p):
print int_p
exit(0)
l_to_r = range(1, len(number)-1)
r_to_l = range(2-len(number), 0)
for i in l_to_r:
if not is_prime(int(number[i:])):
return False
for j in r_to_l:
if not is_prime(int(number[:j])):
return False
return True
mid_len = 1
middles = ['']
result = []
while True:
for middle in middles:
for first in ['2', '3', '5', '7']:
for last in ['3', '7']:
number = first + ''.join(middle) + last
if is_prime(int(number)) and all_prime(number):
result.append(int(number))
if len(result) >= 11:
print result
break
middles = product(['1','3','7','9'], repeat=mid_len)
mid_len += 1
print sum(result)
def main():
acc = 2
for i in range(3, 2000000, 2):
if is_prime(i):
acc += i
print acc
for b in xrange(2, 1000): # b can't be negative
### the following trick could not speed up the program significantly. ###
# trans the formula to: n(n+a)+b
# solve the equotion: n = +-b, n+a = -+b
#possible_n = sorted((-b, b, -b-a, b-a))
#upper_limit = None
#for p_n in possible_n:
# if p_n > 0:
# upper_limit = p_n
# break
#if upper_limit != None and upper_limit <= maximum[0]:
# continue
### end of trick ###
n = 0
while True:
if is_prime(formula(a,b,n)):
n += 1
else:
break
if n > maximum[0]:
maximum = [n, a, b] # n stands for the number of primes. It's a trick.
print a, b, n###
print maximum
