project_euler.problem_027.sol1

Project Euler Problem 27 https://projecteuler.net/problem=27

Problem Statement:

Euler discovered the remarkable quadratic formula: n2 + n + 41 It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 412 + 41 + 41 is clearly divisible by 41. The incredible formula n2 - 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is -126479. Considering quadratics of the form: n² + an + b, where |a| < 1000 and |b| < 1000 where |n| is the modulus/absolute value of ne.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.

Functions

is_prime(→ bool)	Checks to see if a number is a prime in O(sqrt(n)).
solution(→ int)

Module Contents

project_euler.problem_027.sol1.is_prime(number: int) → bool

Checks to see if a number is a prime in O(sqrt(n)). A number is prime if it has exactly two factors: 1 and itself. Returns boolean representing primality of given number num (i.e., if the result is true, then the number is indeed prime else it is not).

>>> is_prime(2)
True
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(2999)
True
>>> is_prime(0)
False
>>> is_prime(1)
False
>>> is_prime(-10)
False

project_euler.problem_027.sol1.solution(a_limit: int = 1000, b_limit: int = 1000) → int

>>> solution(1000, 1000)
-59231
>>> solution(200, 1000)
-59231
>>> solution(200, 200)
-4925
>>> solution(-1000, 1000)
0
>>> solution(-1000, -1000)
0