diff --git a/src/Python/Problems001-050/Problem027.py b/src/Python/Problems001-050/Problem027.py
index 4bb74e1..03caeb3 100644
--- a/src/Python/Problems001-050/Problem027.py
+++ b/src/Python/Problems001-050/Problem027.py
@@ -1,4 +1,4 @@
-#!/usr/bin/env python3
+#!/usr/bin/env python
 """
 Created on 15 Sep 2019
 
@@ -9,7 +9,7 @@ Solution for problem 27 of Project Euler
 https://projecteuler.net/problem=27
 """
 
-from utils import timeit, is_prime
+from utils import is_prime, timeit
 
 
 @timeit("Problem 27")
@@ -24,35 +24,36 @@ def compute():
     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
+    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.
+    -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
+    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
+
+    limit = 1000
     consecutive_values = 0
 
-    for a in range(-999, LIMIT):
-        for b in range(LIMIT + 1):
+    for a in range(-999, limit):
+        for b in range(limit + 1):
             n = 0
-            while is_prime(abs((n ** 2) + (a * n) + b)):
+            while is_prime(abs((n**2) + (a * n) + b)):
                 n += 1
                 if n > consecutive_values:
                     consecutive_values = n
-                    c = a * b
-    return c
+                    ans = a * b
+
+    return ans
 
 
 if __name__ == "__main__":
-
-    print(f"Result for Problem 27: {compute()}")
\ No newline at end of file
+    print(f"Result for Problem 27 is {compute()}")