def run():
sieve = sieve_of_eratosthenes(1000000, lower_limit=10)
# builds masks
bins = {i: [] for i in range(1, 7)}
for i in range(1, 7):
bins[i].extend([
j for j in product(('0', '1'), repeat=i)
if sum(int(k) for k in j) == 3
])
for prime in sieve:
binaries = bins[len(str(prime)) - 1]
for b in binaries:
res = number_primes(prime, b)
if res[0] == 8:
return res[1]
from Utils import sieve_of_eratosthenes
def rotate(s):
return s[len(s) - 1] + s[:len(s) - 1]
sieve = sieve_of_eratosthenes(999999)
table = {k: False for k in range(1000000)}
for i in sieve:
table[i] = True
circular_primes = []
for i in sieve:
if i not in circular_primes:
s = str(i)
rotations = []
for k in s:
s = rotate(s)
rotations.append(int(s))
if not table[int(s)]:
rotations = []
break
circular_primes += rotations
print(len(set(circular_primes)))
# same as 76 only nums is different
from Utils import sieve_of_eratosthenes
limit = 1000
table = {i: 0 for i in range(-limit + 1, limit + 1)}
table[0] = 1
nums = sieve_of_eratosthenes(limit)
for num in nums:
for amount in range(1, limit + 1):
table[amount] += table[amount - num]
for k, v in table.items():
if v > 5000:
print(k)
break
# the less primefactors a number has the closer is the euler-totient to it
# checking every possible number with two factors yields the correct result
from sympy import factorint
from math import sqrt
from Utils import sieve_of_eratosthenes
# returns euler totient given the only two factors
def euler_totient(n, two_factors):
return int(n * (1 - (1 / two_factors[0])) * (1 - (1 / two_factors[1])))
# True if a and b have the same digits
def samedigits(a, b):
return sorted([int(i) for i in str(a)]) == sorted([int(i) for i in str(b)])
limit = 10**7
sieve = sieve_of_eratosthenes(4000)
min = 1000
for i, prime1 in enumerate(sieve):
for j in range(i, len(sieve)):
prime2 = sieve[j]
product = prime1 * prime2
if product > limit:
break
eul = euler_totient(product, (prime1, prime2))
if product / eul < min and samedigits(product, eul):
min = (product) / eul
min_prod = product
print(min_prod)
from Utils import sieve_of_eratosthenes
# no need for euler totient function, because only small numbers
def recur_len(d):
if d % 2 != 0 and d % 5 != 0 and d != 0 and d != 1:
n = 1
# every fraction 1/d can be written as x/(10**n - 1) where n is the length
# of the recurring cycle
while (10**n - 1) % d != 0:
n += 1
return n
res = 0
res_index = 0
sieve = sieve_of_eratosthenes(1000,lower_limit=6)
for i in reversed(sieve):
length = recur_len(i)
if length > res:
res = length
res_index = i
if i < res:
break
print(res_index)
from Utils import sieve_of_eratosthenes print(sum(sieve_of_eratosthenes(2000000)))
from Utils import sieve_of_eratosthenes
def samedigits(a,b):
a = sorted(str(a))
b = sorted(str(b))
return a == b
sieve = sieve_of_eratosthenes(9999,lower_limit=1000)
for i in range(len(sieve)-1):
prime1 = sieve[i]
for j in range(i+1,len(sieve)):
prime2 = sieve[j]
# test if same digits
if samedigits(prime1,prime2):
prime3 = 2*prime2 - prime1
# test if prime3 is prime
if prime3 in sieve and samedigits(prime1,prime3):
concat = str(prime1)+str(prime2)+str(prime3)
if concat != '148748178147':
print(concat)