Euler Problem 27 solution script Added

project_euler/Problem_27/__init__.py

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+

project_euler/Problem_27/sol1.py

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+"""
+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.
+"""
+
+import math
+
+
+def isPrime(k):
+    # checks if a number is prime
+    if k < 2:
+        return False
+    elif k == 2:
+        return True
+    elif k % 2 == 0:
+        return False
+    else:
+        for x in range(3, int(math.sqrt(k) + 1), 2):
+            if k % x == 0:
+                return False
+
+    return True
+
+
+def solution(a_limit, b_limit):
+    """
+	>>> solution(1000, 1000)
+	-59231
+	>>> solution(2000, 2000)
+	-126479
+	>>> solution(-1000, 1000)
+	0
+	>>> solution(-1000, -1000)
+	0
+
+	"""
+    longest = [0, 0, 0]
+    # length, a, b
+    for a in range((a_limit * -1) + 1, a_limit):
+        for b in range(2, b_limit):
+            if isPrime(b):
+                count = 0
+                n = 0
+                while isPrime((n ** 2) + (a * n) + b):
+                    count += 1
+                    n += 1
+
+                if count > longest[0]:
+                    longest = [count, a, b]
+
+    ans = longest[1] * longest[2]
+    return ans
+
+
+if __name__ == "__main__":
+    print(solution(1000, 1000))