def euler0007(n): # n must be 10001 in order to solve the problem
""" Find the 10001st prime. """
P,c = 2,0
while True:
if euler.prime(P):
c+=1
print P,c
if c == n: return P,c
P+=1
def euler0026(n): # n must be 1000 in order to solve the problem
""" Find the value of d < 1000 for which 1/d contains the longest recurring cycle. """
IDC = ''
N = 0
for x in range(1,n):
if euler.prime(x):
idc = (((euler.iDivision(x)).partition('('))[-1].partition(')'))[0]
if len(idc) > len(IDC): IDC,N = idc,x
return IDC,'r',N
def euler0027(n1,n2): # n1 and n2 must be 1000 in order to solve the problem
""" Find a quadratic formula that produces the maximum number of primes for consecutive values of n. """
MAX,R = 0,0
for a in range(-n1,n1):
for b in range(-n2,n2):
n,prime = 0,True
while prime: prime,n = euler.prime(((n*n)+(a*n)+(b))),n+1
n-=1
if n > MAX: MAX,R = n,(a,b,a*b)
# print MAX,n,a,b
return 'the answer is ',R
def main():
nd = 4
# do not limit to len(set(str(p))) == 4
primes = [p for p in prime(10**nd) if p > 10**(nd-1)]
sorted_primes = sorted(primes, key=lambda i: sorted(str(i)))
for k, it in groupby(sorted_primes, key=lambda i: sorted(str(i))):
l = list(it)
length = len(l)
if length < 3:
continue
for i in xrange(1, length):
b = l[i]
# if any(2*b == a+c for a in la for c in lc):
for a in l[:i]:
for c in l[i+1:]:
if 2 * b == a + c and b != 4817:
return int(''.join(map(str, [a, b, c])))
l=[]
l.append(1)
l.append(tri(i))
for j in range(2,int(math.sqrt(tri(i)))+1):
if(not tri(i)%j):
l.append(j)
l.append(tri(i)/j)
if(len(l)>500):
print i
print tri(i)
break
"""
# 3 second implementation
L=[]
for x in euler.prime():
L.append(x)
if len(L) > 2000:
break
i = f = x = 0
while (f <= 500):
i = i + 1
x = x + i
f = 2
a = 1
j = 0
y = x
while (y > 1):
k = 1
while (x%(L[j]**k) == 0):
"""Smallest multiple"""
from euler import prime
N = 20
primes = list(prime(N+1))
for i in range(2, N+1):
prime_factor = []
for p in primes:
while i % p == 0:
prime_factor.append(p)
if i == p:
break
i = i / p
if i == p: break
for each in set(prime_factor):
if prime_factor.count(each) > primes.count(each):
primes.extend([each]*(prime_factor.count(each) - primes.count(each)))
import operator
import functools
print(functools.reduce(operator.mul, primes))
import euler
total = 0
x = 1
euler.primeCache(2000000)
while euler.prime(x) < 2000000:
total += euler.prime(x)
x+=1
print total
def primeProof(x):
xstr = str(x)
for i in range(len(xstr)):
for a in range(10):
ystr = xstr[:i]+str(a)+xstr[i+1:]
if euler.primeQ(int(ystr)): return False
return True
euler.primeCache(200000)
squbes = [2*2*2*q*q for q in euler.primes]
stopper = max(squbes)
for p in xrange(1, len(euler.primes)):
q = 2
while euler.prime(p)**2 * euler.prime(q)**3 < stopper:
if p == q : q +=1
squbes.append(euler.prime(p)**2 * euler.prime(q)**3)
q+=1
squbes.sort()
squbes = filter(twohundred, squbes)
i = 1
for can in squbes:
if primeProof(can):
print can, i
if i == 200:
print can
break
i+=1
import euler print euler.prime(10001)
def primeProof(x):
xstr = str(x)
for i in range(len(xstr)):
for a in range(10):
ystr = xstr[:i] + str(a) + xstr[i + 1:]
if euler.primeQ(int(ystr)): return False
return True
euler.primeCache(200000)
squbes = [2 * 2 * 2 * q * q for q in euler.primes]
stopper = max(squbes)
for p in xrange(1, len(euler.primes)):
q = 2
while euler.prime(p)**2 * euler.prime(q)**3 < stopper:
if p == q: q += 1
squbes.append(euler.prime(p)**2 * euler.prime(q)**3)
q += 1
squbes.sort()
squbes = filter(twohundred, squbes)
i = 1
for can in squbes:
if primeProof(can):
print can, i
if i == 200:
print can
break
i += 1
import euler
p=euler.prime(10000000)
a=[]
for x in p:
if '1' in str(x):
if '2' in str(x):
if '3' in str(x):
if '4' in str(x):
if '5' in str(x):
if '6' in str(x):
if '7' in str(x):a.append(x)
print a[-1]
#!/usr/bin/env python
# By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we
# can see that the 6th prime is 13. What is the 10001st prime number?
import euler
x=0
for prime in euler.prime():
x += 1
if x == 10001:
print prime
break
"""10001st prime"""
from euler import prime
N = 10001
M = 10**6
for index, number in enumerate(prime(M), start=1):
if index == N:
print number
break
def truncatable_prime(n):
for x in xrange(1, len(str(n))):
if not prime(str(n)[x:]) or not prime(str(n)[:x]):
return False
return True
"""Largest prime factor"""
from euler import prime
#M = int(raw_input('Input the integer: '))
#if M < 2: print 'Invalid integer'
M = 600851475143
N = M
# p: prime number no bigger than N
for p in prime(N+1):
while N % p == 0:
if N == p:
print 'Largest prime factor of number {} is: {}'.format(M, p)
raise SystemExit
N = N / p
def circular_prime(n):
for i in xrange(len(str(n))):
if not prime(int(str(n)[i:] + str(n)[:i])):
return False
return True
def euler0003(n): # n must be 600851475143 in order to solve the problem """ Find the largest prime factor of a composite number. """ return ([x for x in euler.mult(n) if euler.prime(x)])
"""
10001st prime
Problem 7
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can
see that the 6th prime is 13.
What is the 10,001st prime number?
"""
from euler import euler_test, prime
# function prime() moved to euler module
if __name__ == "__main__":
euler_test(7, prime(10001))
#!/usr/bin/env python
# The number, 197, is called a circular prime because all rotations of
# the digits: 197, 971, and 719, are themselves prime. There are
# thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71,
# 73, 79, and 97. How many circular primes are there below one
# million?
import euler
for p in euler.prime():
if p > 10**2:#6:
break
primes.add(p)
circular = set()
for p in primes:
import euler
top = 1000
def lth(a, b):
n = 0
while euler.primeQ(n**2 + a * n + b):
n += 1
return n
max = 0
for a in xrange(-1 * top + 1, top, 2): #odds only
b = 1
while euler.prime(b) < top:
p = euler.prime(b)
if lth(a, p) > max:
max, ans = lth(a, p), a * p
b += 1
print ans
l=[]
l.append(1)
l.append(tri(i))
for j in range(2,int(math.sqrt(tri(i)))+1):
if(not tri(i)%j):
l.append(j)
l.append(tri(i)/j)
if(len(l)>500):
print i
print tri(i)
break
"""
# 3 second implementation
L = []
for x in euler.prime():
L.append(x)
if len(L) > 2000:
break
i = f = x = 0
while (f <= 500):
i = i + 1
x = x + i
f = 2
a = 1
j = 0
y = x
while (y > 1):
k = 1
while (x % (L[j]**k) == 0):
"""Summation of primes""" import euler N = 2*10**6 print sum(i for i in euler.prime(N))
import euler
top = 1000
def lth(a,b):
n = 0
while euler.primeQ(n**2 + a*n + b):
n+=1
return n
max = 0
for a in xrange(-1*top+1, top,2): #odds only
b = 1
while euler.prime(b)<top:
p = euler.prime(b)
if lth(a,p) > max:
max, ans = lth(a,p), a*p
b+=1
print ans
def largest_n_digit_pandigital_prime(n):
p = permutations('987654321'[9-n:])
return max([int(''.join(x)) for x in p if prime(''.join(x))])