def goldbach(n):
kontrol = allprimes(n)
for p in kontrol:
if(int(sqrt((n-p)/2))==sqrt((n-p)/2)):
return(1)
break
else:
return(0)
def goldbach(n):
kontrol = allprimes(n)
for p in kontrol:
if (int(sqrt((n - p) / 2)) == sqrt((n - p) / 2)):
return (1)
break
else:
return (0)
# It turns out that the formula will produce 40 primes for the consecutive integer values 0≤n≤390≤n≤39. However, when n=40,40^2+40+41=40(40+1)+41n=40,40^2+40+41=40(40+1)+41 is divisible by 41, and certainly when n=41,41^2+41+41n=41,41^2+41+41 is clearly divisible by 41.
# The incredible formula n^2−79n+1601 was discovered, which produces 80 primes for the consecutive values 0≤n≤790≤n≤79. The product of the coefficients, −79 and 1601, is −126479.
# Considering quadratics of the form:
# n^2+an+b, where |a|<1000|a|<1000 and |b|≤1000|b|≤1000
# where |n||n| is the modulus/absolute value of nn
# e.g. |11|=11|11|=11 and |−4|=4|−4|=4
# Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n=0.
from functions import allprimes
from functions import isprime
def quadratic(a,b):
primecount = 0
for n in range(100):
if(isprime(n**2+a*n+b)==1):
primecount += 1
else:
break
return(primecount)
blist = allprimes(1000)
maxab = 1
sonuc = None
for a in range(-999,1000):
for b in blist:
if(quadratic(a,b)>maxab):
maxab = quadratic(a,b)
sonuc = a*b
print(sonuc)
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# https://projecteuler.net/problem=266
from functions import allprimes
from math import sqrt
from bisect import bisect_left
primes = allprimes(190)
p = 1
for n in primes:
p = p * n
kok = sqrt(p)
def euler(n):
nrs = {1}
for m in primes[n::2]:
nrs |= {nr * m for nr in nrs}
return (sorted(nrs))
nrs1 = euler(0)
nrs2 = euler(1)
maxpsqr = 0
for m in nrs1:
psqr = nrs2[bisect_left(nrs2, kok / m) - 1] * m
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# 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?
from functions import allprimes
primes = allprimes(10000)[168:]
sonuc = 0
for n in primes:
for m in range(2, 5000 - int(n / 2), 2):
if (primes.count(n + m) == 1 and primes.count(n + 2 * m) == 1):
for i in str(n):
if (str(n + m).count(i) == 0 or str(n + 2 * m).count(i) == 0):
break
else:
for j in str(n + m):
if (str(n + 2 * m).count(j) == 0 or str(n).count(j) == 0):
break
else:
for k in str(n + 2 * m):
if (str(n).count(k) == 0 or str(n + m).count(k) == 0):
break
else:
print(n, n + m, n + 2 * m)
sonuc += 1
if (sonuc == 2):
break
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
# Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
# NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
from functions import allprimes
primes = allprimes(750000)
rightprimes = []
for n in primes[4:]:
temp = n
n = list(str(n))
for i in range(1, len(n)):
n.pop()
n = ''.join(str(i) for i in n)
if (primes.count(int(n)) == 0):
break
n = list(str(n))
else:
rightprimes.append(temp)
leftprimes = []
for n in rightprimes:
temp = n
n = list(str(n))
for i in range(1, len(n)):
n.remove(n[0])
n = ''.join(str(i) for i in n)
if (primes.count(int(n)) == 0):
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# For a prime p let S(p) = (∑(p-k)!) mod(p) for 1 ≤ k ≤ 5.
# For example, if p=7,
# (7-1)! + (7-2)! + (7-3)! + (7-4)! + (7-5)! = 6! + 5! + 4! + 3! + 2! = 720+120+24+6+2 = 872.
# As 872 mod(7) = 4, S(7) = 4.
# It can be verified that ∑S(p) = 480 for 5 ≤ p < 100.
# Find ∑S(p) for 5 ≤ p < 108.
# Soruda verilen S fonksiyonu (p-5)!+(p-4)!+(p-3)!+(p-2)!+(p-1)! (mod p) şeklinde yazılabilir.
# Bu ifadeyi (p-1)! parantezine alıp (p-2)(mod p) yerine -2 yazıldığında
# (elbette diğerleri için de aynı şeyi yaparak) ve Wilson Teoremini kullanarak S(p)=-3/8(mod p) sonucuna ulaşılır.
# Buradan da S(p)=(p-3)/8 (mod p) elde edilir.
# Kolayca S(p)'nin ancak p'nin asal olduğu durumlarda sıfırdan farklı değer verdiği görülebilir.
# p'nin asal olduğu bilinince Fermatın Küçük Teoremi kullanılarak paydadaki 8'in tersi bulunabilir.
from functions import allprimes
def s(p):
return(((p-3)*pow(8,p-2,p))%p)
primes = allprimes(10**8)[2:]
cozum = 0
for p in primes:
cozum += s(p)
print(cozum)
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
# Let us consider repunits of the form R(10^n).
# Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is divisible by 17. Yet there is no value of n for which R(10^n) will divide by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of R(10^n).
# Find the sum of all the primes below one-hundred thousand that will never be a factor of R(10^n).
# https://en.wikipedia.org/wiki/Repunit adresinden de görülebileceği gibi bu sayılar R(n)=10^n-1 şeklinde yazılabiliyor.
from functions import allprimes
primes = allprimes(100000)
bolen = 3
L = 10**20
for n in primes:
if(pow(10, L, n) != 1):
bolen += n
print(bolen)
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# Consider the divisors of 30: 1,2,3,5,6,10,15,30.
# It can be seen that for every divisor d of 30, d+30/d is prime.
# Find the sum of all positive integers n not exceeding 100 000 000 such that for every divisor d of n, d+n/d is prime.
from functions import allprimes
LIMIT = 10**8
primes = allprimes(LIMIT)
def kontrol(n):
if n % 4 == 0: return 0
r = int(n**0.5) + 1
for i in range(1, r + 1):
if (n % i == 0):
if i + n / i not in s:
return 0
return 1
s = set(primes)
toplam = sum(p - 1 for p in primes if kontrol(p - 1))
print(toplam)
from functions import allprimes
from functions import isprime
import random
deneme = 0
koltuklar=allprimes(100)
while(len(koltuklar)>0):
tercih = random.randint(1,200)%98
deneme += 1
if(isprime(tercih)==1 and koltuklar.count(tercih)==1):
koltuklar.remove(tercih)
print(deneme)
i, n = 2, m
while (i <= m / 2):
if n % i == 0:
carpanlar.append(i)
n = n / i
else:
i += 1
sonuc = []
while not (len(carpanlar) == 0):
a = carpanlar.count(carpanlar[0])
sonuc.append(carpanlar[0]**a)
carpanlar = carpanlar[a:]
return (sonuc)
asal = allprimes(200000)
for n in range(130000, 200000):
if (asal.count(n) == 1):
continue
a = primedivisors(n)
if (asal.count(n + 1) == 1):
continue
b = primedivisors(n + 1)
if (asal.count(n + 2) == 1):
continue
c = primedivisors(n + 2)
if (asal.count(n + 3) == 1):
continue
d = primedivisors(n + 3)
if (len(a) != 4 or len(b) != 4 or len(c) != 4 or len(d) != 4):
continue
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# 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?
from functions import allprimes
primes=allprimes(10000)[168:]
sonuc=0
for n in primes:
for m in range(2,5000-int(n/2),2):
if(primes.count(n+m)==1 and primes.count(n+2*m)==1):
for i in str(n):
if(str(n+m).count(i)==0 or str(n+2*m).count(i)==0):
break
else:
for j in str(n+m):
if(str(n+2*m).count(j)==0 or str(n).count(j)==0):
break
else:
for k in str(n+2*m):
if(str(n).count(k)==0 or str(n+m).count(k)==0):
break
else:
print(n,n+m,n+2*m)
sonuc += 1
if(sonuc==2):
break
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# 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 functions import allprimes
from functions import ispandigital
primes=allprimes(10**7)[::-1]
for n in primes:
if(ispandigital(n)==1):
print(n)
break
# It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.
# 9 = 7 + 2×1^2
# 15 = 7 + 2×2^2
# 21 = 3 + 2×3^2
# 25 = 7 + 2×3^2
# 27 = 19 + 2×2^2
# 33 = 31 + 2×1^2
# It turns out that the conjecture was false.
# What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
from functions import allprimes
from math import sqrt
def goldbach(n):
kontrol = allprimes(n)
for p in kontrol:
if(int(sqrt((n-p)/2))==sqrt((n-p)/2)):
return(1)
break
else:
return(0)
asal = allprimes(10000)
for n in range(9,10000,2):
if(asal.count(n)==1):
continue
if(goldbach(n)==0):
print(n)
break
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# What is the largest prime factor of the number 600851475143 ?
from functions import allprimes
primes=allprimes(10000)
number,i=600851475143,1
while not (number%primes[-i]==0):
i +=1
print(primes[-i])
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
# Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
# NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
from functions import allprimes
primes=allprimes(750000)
rightprimes = []
for n in primes[4:]:
temp = n
n=list(str(n))
for i in range(1,len(n)):
n.pop()
n=''.join(str(i) for i in n)
if(primes.count(int(n))==0):
break
n=list(str(n))
else:
rightprimes.append(temp)
leftprimes =[]
for n in rightprimes:
temp = n
n=list(str(n))
for i in range(1,len(n)):
n.remove(n[0])
n=''.join(str(i) for i in n)
if(primes.count(int(n))==0):
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# The prime 41, can be written as the sum of six consecutive primes:
# 41 = 2 + 3 + 5 + 7 + 11 + 13
# This is the longest sum of consecutive primes that adds to a prime below one-hundred.
# The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
# Which prime, below one-million, can be written as the sum of the most consecutive primes?
from functions import isprime
from functions import allprimes
asal = allprimes(4000)
toplam = []
for p in range(5,len(asal)):
for a in range(0,len(asal)-p):
if(isprime(sum(asal[a:a+p]))==1 and sum(asal[a:a+p])<1000000):
toplam.append(sum(asal[a:a+p]))
print(toplam[-1])
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# Consider the divisors of 30: 1,2,3,5,6,10,15,30.
# It can be seen that for every divisor d of 30, d+30/d is prime.
# Find the sum of all positive integers n not exceeding 100 000 000 such that for every divisor d of n, d+n/d is prime.
from functions import allprimes
LIMIT = 10**8
primes = allprimes(LIMIT)
def kontrol(n):
if n%4==0 : return 0
r=int(n**0.5)+1
for i in range(1,r+1):
if(n%i==0):
if i+n/i not in s:
return 0
return 1
s=set(primes)
toplam=sum(p-1 for p in primes if kontrol(p-1))
print(toplam)
carpanlar = []
i,n=2,m
while (i<=m/2):
if n%i==0:
carpanlar.append(i)
n = n/i
else:
i += 1
sonuc = []
while not(len(carpanlar)==0):
a=carpanlar.count(carpanlar[0])
sonuc.append(carpanlar[0]**a)
carpanlar = carpanlar[a:]
return(sonuc)
asal = allprimes(200000)
for n in range(130000,200000):
if(asal.count(n)==1):
continue
a = primedivisors(n)
if(asal.count(n+1)==1):
continue
b = primedivisors(n+1)
if(asal.count(n+2)==1):
continue
c = primedivisors(n+2)
if(asal.count(n+3)==1):
continue
d = primedivisors(n+3)
if(len(a)!=4 or len(b)!=4 or len(c)!=4 or len(d)!=4):
continue
#!/usr/bin/python3 # -*- coding: utf-8 -*- # Find the sum of all the primes below two million. from functions import allprimes print(sum(allprimes(2000000)))
from functions import allprimes
from functions import isprime
import random
deneme = 0
koltuklar = allprimes(100)
while (len(koltuklar) > 0):
tercih = random.randint(1, 200) % 98
deneme += 1
if (isprime(tercih) == 1 and koltuklar.count(tercih) == 1):
koltuklar.remove(tercih)
print(deneme)
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# 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 functions import allprimes
from functions import ispandigital
primes=allprimes(10000000)[::-1]
for n in primes:
if(ispandigital(n)==1):
print(n)
break
# 15 = 7 + 2×2^2
# 21 = 3 + 2×3^2
# 25 = 7 + 2×3^2
# 27 = 19 + 2×2^2
# 33 = 31 + 2×1^2
# It turns out that the conjecture was false.
# What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
from functions import allprimes
from math import sqrt
def goldbach(n):
kontrol = allprimes(n)
for p in kontrol:
if (int(sqrt((n - p) / 2)) == sqrt((n - p) / 2)):
return (1)
break
else:
return (0)
asal = allprimes(10000)
for n in range(9, 10000, 2):
if (asal.count(n) == 1):
continue
if (goldbach(n) == 0):
print(n)
break
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# For a prime p let S(p) = (∑(p-k)!) mod(p) for 1 ≤ k ≤ 5.
# For example, if p=7,
# (7-1)! + (7-2)! + (7-3)! + (7-4)! + (7-5)! = 6! + 5! + 4! + 3! + 2! = 720+120+24+6+2 = 872.
# As 872 mod(7) = 4, S(7) = 4.
# It can be verified that ∑S(p) = 480 for 5 ≤ p < 100.
# Find ∑S(p) for 5 ≤ p < 108.
# Soruda verilen S fonksiyonu (p-5)!+(p-4)!+(p-3)!+(p-2)!+(p-1)! (mod p) şeklinde yazılabilir.
# Bu ifadeyi (p-1)! parantezine alıp (p-2)(mod p) yerine -2 yazıldığında
# (elbette diğerleri için de aynı şeyi yaparak) ve Wilson Teoremini kullanarak S(p)=-3/8(mod p) sonucuna ulaşılır.
# Buradan da S(p)=(p-3)/8 (mod p) elde edilir.
# Kolayca S(p)'nin ancak p'nin asal olduğu durumlarda sıfırdan farklı değer verdiği görülebilir.
# p'nin asal olduğu bilinince Fermatın Küçük Teoremi kullanılarak paydadaki 8'in tersi bulunabilir.
from functions import allprimes
def s(p):
return (((p - 3) * pow(8, p - 2, p)) % p)
primes = allprimes(10**8)[2:]
cozum = 0
for p in primes:
cozum += s(p)
print(cozum)
#!/usr/bin/python3
# -*- coding: utf-8 -*-
# A composite is a number containing at least two prime factors. For example, 15 = 3 × 5; 9 = 3 × 3; 12 = 2 × 2 × 3.
# There are ten composites below thirty containing precisely two, not necessarily distinct, prime factors: 4, 6, 9, 10, 14, 15, 21, 22, 25, 26.
# How many composite integers, n < 10^8, have precisely two, not necessarily distinct, prime factors?
# 10^8'den yarısından küçük her asal için kendinden küçük eşit asal sayılarla çarpımından oluşturulan sayılar sayıldığında aranan sonuç bulunmuş oluyor.
from sympy import primepi
from functions import allprimes
sonuc = 0
LIMIT = 10**8
primes = allprimes(LIMIT // 2)
for n in primes:
if (n < LIMIT**(0.5)):
m = primepi(n)
else:
m = primepi(LIMIT // n)
sonuc += m
print(sonuc)
