Solution to problem 27

src/Python/Problem027.py

@@ -0,0 +1,58 @@
+#!/usr/bin/env python3
+"""
+Created on 15 Sep 2019
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for problem 27 of Project Euler
+https://projecteuler.net/problem=27
+"""
+
+from utils import timeit, is_prime
+
+
+@timeit("Problem 27")
+def compute():
+    """
+    Euler discovered the remarkable quadratic formula:
+
+        n^2 + n + 41
+
+    It turns out that the formula will produce 40 primes for the consecutive
+    integer values 0≤n≤39. However, when n=40, 40^2+40+41=40(40+1)+41 is
+    divisible by 41, and certainly when n=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≤79. The product of the coefficients,
+    −79 and 1601, is −126479.
+
+    Considering quadratics of the form:
+
+        n^2 + an + b
+
+    where |a|<1000, |b|≤1000 and |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.
+    """
+    LIMIT = 1000
+    consecutive_values = 0
+
+    for a in range(-999, LIMIT):
+        for b in range(LIMIT + 1):
+            n = 0
+            while is_prime(abs((n ** 2) + (a * n) + b)):
+                n += 1
+                if n > consecutive_values:
+                    consecutive_values = n
+                    c = a * b
+    return c
+
+
+if __name__ == "__main__":
+
+    print(f"Result for Problem 27: {compute()}")