from utils.prime import is_prime
ans = 0
for i in range(2, 2000000):
if is_prime(i):
ans += i
print(ans)
# -*- coding: utf-8 -*-
"""
Problem 41 - Pandigital prime
=============================
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 utils.number import gen_permutation
from utils.prime import is_prime
digits = [9, 8, 7, 6, 5, 4, 3, 2, 1]
found = False
i = 0
while not found and i < len(digits):
for permutation in gen_permutation(digits[i:]):
n = int(''.join(map(str, permutation)))
if is_prime(n):
found = True
print(n)
break
i += 1
# Answer: 7652413
where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |-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 utils.prime import is_prime, gen_prime
primes = []
for prime in gen_prime():
if prime > 1000:
break
primes.append(prime)
max_count = 0
product = 0
for a in range(-999, 1000):
for b in primes:
n = 0
while is_prime(n**2 + a * n + b):
n += 1
if n > max_count:
max_count = n
product = a * b
print(product)
# Answer: -59231
def check(num):
x = int(math.sqrt(num // 2)) # x is the bound for square number
for i in range(1, x + 1):
if is_prime(num - 2 * i * i):
return True
return False
# in the name of God
import math
from utils.prime import is_prime
def check(num):
x = int(math.sqrt(num // 2)) # x is the bound for square number
for i in range(1, x + 1):
if is_prime(num - 2 * i * i):
return True
return False
for i in range(3, 10000000, 2):
if not is_prime(i) and not check(i):
print(i)
break
def test_is_prime(self, n: int, expected: bool) -> None:
self.assertEqual(is_prime(n), expected)
import math
from utils.prime import is_prime
pivot = 600851475143
ans = 0
for i in range(2, int(math.sqrt(pivot)) + 10):
if pivot % i != 0:
continue
if is_prime(i):
ans = max(ans, i)
if is_prime(pivot // i):
ans = max(ans, pivot // i)
print(ans)
from utils.prime import is_prime
primes = []
for i in range(2, 10 ** 6):
if is_prime(i):
primes.append(i)
# calculate partial sum for consecutive primes
partial_sum = [primes[0]]
for i in range(1, len(primes)):
partial_sum.append(partial_sum[i - 1] + primes[i])
maxi = 0
ans = -1
for i in range(len(primes)):
for j in range(i + 1, min(i + 1000, len(primes))): # do you know why i + 1000? ask if u cant get the meaning
primes_sum = partial_sum[j] - (partial_sum[i - 1] if i > 0 else 0)
if primes_sum < 10 ** 6 and j - i + 1 > maxi and is_prime(primes_sum):
maxi = j - i + 1
ans = primes_sum
print(maxi, ans)