diff --git a/e026.py b/e026.py
new file mode 100644
index 0000000..b69c726
--- /dev/null
+++ b/e026.py
@@ -0,0 +1,97 @@
+# -*- coding: utf-8 -*-
+"""Find the value of d < 1000 for which 1/d contains the longest recurring cycle.A unit fraction contains 1 in the numerator.
+
+The decimal representation of the unit fractions with denominators 2 to 10 are given:
+    1/2	= 	0.5
+    1/3	= 	0.(3)
+    1/4	= 	0.25
+    1/5	= 	0.2
+    1/6	= 	0.1(6)
+    1/7	= 	0.(142857)
+    1/8	= 	0.125
+    1/9	= 	0.(1)
+    1/10	= 	0.1
+Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
+
+Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
+"""
+
+from decimal import Decimal
+from e003 import pfactor
+
+def coprime(a, b):
+    """Are integers a and b coprime?"""
+
+    primes_a = pfactor(a)
+    primes_b = pfactor(b)
+    primes = [p for p in primes_a if p in primes_b]
+    return bool(primes)
+
+def long_division(dividend, divisor):
+    dividend = str(dividend).replace('.', '')
+    i = -1
+    remainder = 0
+    while True:
+        while remainder < divisor:
+            i += 1
+            remainder *= 10
+            try:
+                remainder += int(dividend[i])
+            except:
+                pass
+            if remainder == 0:
+                return
+        digit = str(remainder // divisor)
+        remainder = remainder % divisor
+        yield digit, remainder
+
+def get_cycle_length(numerator, denominator):
+    """Determine the period length of a non-terminating decimal"""
+
+    length = None
+    numerator = Decimal(numerator)
+    denominator = Decimal(denominator)
+
+    if denominator == 1:
+        return 1
+    
+    """
+    From Wikipedia:
+        Terminating decimals represent rational numbers of the form
+        k/2^n5^m
+
+        However, a terminating decimal also has a representation as a
+        repeating decimal, obtained by decreasing the final (nonzero)
+        digit by one and appending an infinitely repeating sequence of
+        nines. 1 = 0.999999... and 1.585 = 1.584999999... are two
+        examples of this.
+    """
+    if coprime(denominator, 10):
+        return 1
+
+    digits = []
+    for digit in long_division(numerator, denominator):
+        if digit in digits:
+            cycle = digits[digits.index(digit):]
+            return len(cycle)
+        digits.append(digit)
+
+    else:
+        raise Exception('Non-reciprocal period detection not yet implemented')
+    return length
+
+def main():
+    MAX = 1000
+    longest_cycle = (0, 0)
+    for i in xrange((MAX), 1, -1):
+        # The period of 1/k for integer k is always <= k - 1
+        if longest_cycle[1] >= i:
+            break
+        
+        cycle = get_cycle_length(1, i)
+        if longest_cycle[1] < cycle:
+            longest_cycle = (i, cycle)
+    (i, cycle) = longest_cycle
+    print 'Longest cycle length is', cycle, 'for i =', i
+if __name__ == '__main__':
+    main()
diff --git a/e027.py b/e027.py
deleted file mode 100644
index 4d4ae74..0000000
--- a/e027.py
+++ /dev/null
@@ -1,73 +0,0 @@
-# -*- coding: utf-8 -*-
-"""Find the value of d < 1000 for which 1/d contains the longest recurring cycle.A unit fraction contains 1 in the numerator.
-
-The decimal representation of the unit fractions with denominators 2 to 10 are given:
-    1/2	= 	0.5
-    1/3	= 	0.(3)
-    1/4	= 	0.25
-    1/5	= 	0.2
-    1/6	= 	0.1(6)
-    1/7	= 	0.(142857)
-    1/8	= 	0.125
-    1/9	= 	0.(1)
-    1/10	= 	0.1
-Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
-
-Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
-"""
-
-from decimal import Decimal
-from e003 import pfactor
-
-def coprime(a, b):
-    """Are integers a and b coprime?"""
-
-    primes_a = pfactor(a)
-    primes_b = pfactor(b)
-    primes = [p for p in primes_a if p in primes_b]
-    return bool(primes)
-
-def period_length(numerator, denominator):
-    """Determine the period length of a non-terminating decimal"""
-    
-    length = None
-    numerator = Decimal(numerator)
-    denominator = Decimal(denominator)
-    if numerator == 1:
-        if denominator == 1:
-            return 1
-        primes = pfactor(int(denominator))
-        
-        """
-        From Wikipedia:
-            Terminating decimals represent rational numbers of the form
-            k/2^n5^m
-
-            However, a terminating decimal also has a representation as a
-            repeating decimal, obtained by decreasing the final (nonzero)
-            digit by one and appending an infinitely repeating sequence of
-            nines. 1 = 0.999999... and 1.585 = 1.584999999... are two
-            examples of this.
-        """
-        if len([p for p in primes if p not in [2, 5]]) == 0:
-            return 1
-
-        if coprime(denominator, 10):
-            pass
-    else:
-        raise Exception('Non-reciprocal period detection not yet implemented')
-    return length
-
-def main():
-    MAX = 10
-    longest_cycle = 0
-    for i in xrange((MAX), 1, -1):
-        # The period of 1/k for integer k is always ² k ? 1.
-        if longest_cycle >= i:
-            break
-        
-        cycle = period_length(1, i)
-        fraction = Decimal(1) / Decimal(i)
-        print '1/{0} = {1} ({2})'.format(i, fraction, cycle), fraction.is_subnormal()
-if __name__ == '__main__':
-    main()