def isTruncPrime(n):
s = str(n)
pre = ['1', '9']
if s[0] in pre or s[-1] in pre:
return False
for i in range(1, len(s)):
if not isPrime(int(s[:i])) or not isPrime(int(s[i:])):
return False
return True
def isCircularPrime(n):
if '5' in str(n):
return False
if not isPrime(n):
return False
nums = Rotation(str(n))
for n in nums[1:]:
if not isPrime(int(n)):
return False
return True
def run(n):
t = 2
arr = []
while not common.isPrime(n):
if t == 2 and n % t != 0:
t += 1
elif n % t != 0:
t += 2
if not common.isPrime(t):
t += 2
if n % t == 0:
n //= t
arr.append(t)
arr.append(n)
return max(arr)
def isTruncatablePrime(n, leftToRight): s = str(n) while len(s) > 0: if not isPrime(int(s)): return False s = s[1:] if leftToRight else s[:-1] return True
def isCircularPrime(n): s = str(n) for i in range(len(s)): if not isPrime(int(s)): return False s = s[1:] + s[0] return True
def prob10(): maxNumber = 2000000 primeList = set() for i in range(2,maxNumber+1): if isPrime(i): primeList.add(i) return sum(primeList)
def prob7(): primeList = set() i = 2 while len(primeList) <= 10000: if isPrime(i): primeList.add(i) i = i+1 return max(primeList)
def prob3(): number = 600851475143 primeList = set() primeFactors = set() for i in range(2,int(ceil(sqrt(number)))+1): if i in primeList or isPrime(i): primeList.add(i) if number % i == 0: primeFactors.add(i) return max(primeFactors)
def quadratic(a, b):
"""
If n^2 + a*n + b is a quadratic formula then return the number of prime
it produces, else return 0
"""
n = 1
while True:
fn = n * (n + a) + b
if not isPrime(fn):
break
else:
n += 1
return n - 1
# The prime factors of 13195 are 5, 7, 13 and 29. # What is the largest prime factor of the number 600851475143? from math import sqrt from common import isPrime val = 600851475143 if isPrime(val): print(val) exit() for n in range(int(sqrt(val))+1, 0, -1): if val % n == 0 and isPrime(n): print(n) break
If n^2 + a*n + b is a quadratic formula then return the number of prime
it produces, else return 0
"""
n = 1
while True:
fn = n * (n + a) + b
if not isPrime(fn):
break
else:
n += 1
return n - 1
if __name__ == "__main__":
nums, product = -1, 0
primes = [n for n in range(2, 1000) if isPrime(n)]
# derivation of f(n)=n^2+a*n+b is f'=2*n+a, so f(n) get its minimum value when
# n=-a/2, and f(-a/2)=b-(a^2)/4 must be greater than 2(the smallest prime)
# when a is lesser than zero.
# For b is up to 999,so we get 4*999-8 >= a^2, then a >= -63 when a is
# negative, so a starts at -63
for a in range(-63, 1000):
# b must prime because f(n) = b (when n=0)
for b in primes:
if a == 0 or (a < 0 and 4 * b - a ** 2 - 8 < 0):
continue
tmp = quadratic(a, b)
if nums < tmp:
nums = tmp
product = a * b
# The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. # Find the sum of all the primes below two million. from common import isPrime print(2 + sum([n for n in range(3, 2000000+1, 2) if isPrime(n)]))
return s.find(substring)
def getAllPrimes(u):
pdict = dict()
allps = []
for i in range(2, u):
n, ps, flag = i, [], True
while n not in pset:
if n in pdict:
ps += pdict[n]
flag = False
break
else:
for p in primes:
if not n % p:
ps.append(p)
n /= p
break
if flag:
ps.append(n)
allps.append(ps)
pdict[i] = ps
return allps
primes = [i for i in range(2, 1000000) if isPrime(i)]
pset = Set(primes)
distincts = [len(Set(ps)) for ps in getAllPrimes(1000000)]
print findFirsNumber(distincts, '4444') + 2
#! /usr/bin/python2.7
# -*- coding: utf-8 -*-
from common import isPrime
import itertools
# sum(range(1,10)) = 45, which can be divided by 3, so 9-bit number is impossible
# so does 8-bit numbers, while 7-bit numbers might be fit
# starting from the largest 7-bit number, so it doesn't need to check the
# maximum
nums = itertools.permutations(range(7, 0, -1))
for i in nums:
cur = int(''.join([str(item) for item in i]))
if isPrime(cur):
print cur
break
from common import isPrime
cnt = 0
i = 1
while cnt < 10001:
i += 1
if isPrime(i):
cnt += 1
print(i)
#! /usr/bin/python2.7
# -*- coding: utf-8 -*-
from common import isPrime
def isTruncPrime(n):
s = str(n)
pre = ['1', '9']
if s[0] in pre or s[-1] in pre:
return False
for i in range(1, len(s)):
if not isPrime(int(s[:i])) or not isPrime(int(s[i:])):
return False
return True
def isPrimeWithOdd(n):
even = ['0', '4', '6', '8']
for i in even:
if i in str(n):
return False
return True
primes = [i for i in range(11, pow(10, 6)) if isPrime(i) and isPrimeWithOdd(i)]
if __name__ == "__main__":
res = [p for p in primes if isTruncPrime(p)]
print sum(res)
def isPrime(n):
if n < 0:
return False
elif n < len(p):
return p[n]
return common.isPrime(n)
#! /usr/bin/python2.7
# -*- coding: utf-8 -*-
from common import isPrime
from sets import Set
def isConjecture(n):
root = 1
while 2 * pow(root, 2) <= n - 2:
if n - 2 * pow(root, 2) in primes:
return True
root += 1
return False
primes = Set([i for i in range(2, 10000) if isPrime(i)])
odds = Set(range(3, 10000, 2))
oddcomps = odds.difference(primes)
for n in oddcomps:
if not isConjecture(n):
print n
break
# 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? from sys import maxsize from common import isPrime primesToFind = 10001 primesFound = 0 lastPrime = 0 for n in range(2, maxsize): if isPrime(n): lastPrime = n primesFound += 1 if primesFound == primesToFind: break print(lastPrime)
#! /usr/bin/python2.7
# -*- coding: utf-8 -*-
from common import isPrime
# let a, b, c be the increasing sequence, then a + c = 2 * b
# a, b, c all end with the same number like 1, 3, 7, 9
primes = [i for i in range(1000, 9999) if isPrime(i)]
pdict = dict()
pdict['1'], pdict['3'], pdict['7'], pdict['9'] = [], [], [], []
for p in primes:
pdict[str(p)[-1]].append(p)
def findPP(arr):
l = len(arr)
for i in range(l):
for j in range(i + 1, l):
mid = (arr[i] + arr[j]) / 2
if mid in arr and sorted(str(arr[i])) == sorted(str(mid)) == sorted(str(arr[j])):
print str(arr[i]) + str(mid) + str(arr[j])
for k in pdict.keys():
findPP(pdict[k])
