diff --git a/C/p069.c b/C/p069.c
index a55cc23..8f201e5 100644
--- a/C/p069.c
+++ b/C/p069.c
@@ -2,15 +2,15 @@
  * relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
  *
  * n	Relatively Prime	φ(n)	n/φ(n)
- * 2	1	               1	   2
- * 3	1,2	            2	   1.5
- * 4	1,3	            2	   2
- * 5	1,2,3,4	         4	   1.25
- * 6	1,5	            2	   3
- * 7	1,2,3,4,5,6	      6	   1.1666...
- * 8	1,3,5,7	         4	   2
- * 9	1,2,4,5,7,8	      6	   1.5
- * 10	1,3,7,9	         4	   2.5
+ * 2	1	                1	   2
+ * 3	1,2	                2 	   1.5
+ * 4	1,3	                2	   2
+ * 5	1,2,3,4	            4	   1.25
+ * 6	1,5	                2	   3
+ * 7	1,2,3,4,5,6	        6	   1.1666...
+ * 8	1,3,5,7	            4	   2
+ * 9	1,2,4,5,7,8	        6	   1.5
+ * 10	1,3,7,9	            4	   2.5
  *
  * It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
  *
@@ -26,41 +26,41 @@
 
 int main(int argc, char **argv)
 {
-	int i, res = 1;
+    int i, res = 1;
 	double elapsed;
-   struct timespec start, end;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   i = 1;
+    i = 1;
 
-   /* Using Euler's formula, phi(n)=n*prod(1-1/p), where p are the distinct
-    * primes that divide n. So n/phi(n)=1/prod(1-1/p). To find the maximum
-    * value of this function, the denominator must be minimized. This happens
-    * when n has the most distinct small prime factor, i.e. to find the solution
-    * we need to multiply the smallest consecutive primes until the result is
-    * larger than 1000000.*/
-	while(res < N)
-   {
-      i++;
-      
-      if(is_prime(i))
-      {
-         res *= i;
-      }
-	}
+    /* Using Euler's formula, phi(n)=n*prod(1-1/p), where p are the distinct
+     * primes that divide n. So n/phi(n)=1/prod(1-1/p). To find the maximum
+     * value of this function, the denominator must be minimized. This happens
+     * when n has the most distinct small prime factor, i.e. to find the solution
+     * we need to multiply the smallest consecutive primes until the result is
+     * larger than 1000000.*/
+    while(res < N)
+    {
+        i++;
+        
+        if(is_prime(i))
+        {
+            res *= i;
+        }
+    }
 
-   /* We need the previous value, because we want i<1000000.*/
-   res /= i;
+    /* We need the previous value, because we want i<1000000.*/
+    res /= i;
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec-start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec-start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 69\n");
+    printf("Project Euler, Problem 69\n");
 	printf("Answer: %d\n", res);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
 	return 0;
 }
diff --git a/Python/p022.py b/Python/p022.py
index 9965a67..272c9f5 100644
--- a/Python/p022.py
+++ b/Python/p022.py
@@ -27,7 +27,7 @@ def p022():
     sum_ = 0
     i = 1
 
-#   Calculate the score of each name an multiply by its position.
+    # Calculate the score of each name an multiply by its position.
     for name in names:
         l = len(name)
         score = 0
diff --git a/Python/p027.py b/Python/p027.py
index cb2ae22..1f6e578 100644
--- a/Python/p027.py
+++ b/Python/p027.py
@@ -17,7 +17,8 @@
 # where |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.
+# 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.
 
 from projecteuler import is_prime, timing
 
diff --git a/Python/p045.py b/Python/p045.py
index 867cef6..2f827be 100644
--- a/Python/p045.py
+++ b/Python/p045.py
@@ -2,8 +2,8 @@
 
 # Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
 #
-# Triangle	 	T_n=n(n+1)/2	 	1, 3, 6, 10, 15, ...
-# Pentagonal	 	P_n=n(3n−1)/2	        1, 5, 12, 22, 35, ...
+# Triangle	 	T_n=n(n+1)/2	1, 3, 6, 10, 15, ...
+# Pentagonal	P_n=n(3n−1)/2	1, 5, 12, 22, 35, ...
 # Hexagonal	 	H_n=n(2n−1)	 	1, 6, 15, 28, 45, ...
 #
 # It can be verified that T_285 = P_165 = H_143 = 40755.
diff --git a/Python/p054.py b/Python/p054.py
index 7477a5e..a1155ca 100644
--- a/Python/p054.py
+++ b/Python/p054.py
@@ -15,7 +15,7 @@
 # The cards are valued in the order:
 # 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.
 #
-# If two players have the same ranked hands then the rank made up of the highest value wins; for example, a pair of eights beats a pair of fives 
+# If two players have the same ranked hands then the rank made up of the highest value wins; for example, a pair of eights beats a pair of fives
 # (see example 1 below). But if two ranks tie, for example, both players have a pair of queens, then highest cards in each hand are compared
 # (see example 4 below); if the highest cards tie then the next highest cards are compared, and so on.
 #
diff --git a/Python/p055.py b/Python/p055.py
index 773af9a..c7f6953 100644
--- a/Python/p055.py
+++ b/Python/p055.py
@@ -11,8 +11,8 @@
 # That is, 349 took three iterations to arrive at a palindrome.
 #
 # Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome
-# through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, 
-# we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either 
+# through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem,
+# we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either
 # (i) become a palindrome in less than fifty iterations, or,
 # (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown
 # to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).
diff --git a/Python/p062.py b/Python/p062.py
index b62d6f0..e35625d 100644
--- a/Python/p062.py
+++ b/Python/p062.py
@@ -1,6 +1,6 @@
 #!/usr/bin/python
 
-# The cube, 41063625 (345^3), can be permuted to produce two other cubes: 56623104 (384^3) and 66430125 (405^3). 
+# The cube, 41063625 (345^3), can be permuted to produce two other cubes: 56623104 (384^3) and 66430125 (405^3).
 # In fact, 41063625 is the smallest cube which has exactly three permutations of its digits which are also cube.
 #
 # Find the smallest cube for which exactly five permutations of its digits are cube.
diff --git a/Python/p069.py b/Python/p069.py
index f540a09..95b72b4 100644
--- a/Python/p069.py
+++ b/Python/p069.py
@@ -3,15 +3,15 @@
 # Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of numbers less than n which are
 # relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
 #
-# n	Relatively Prime    φ(n)    n/φ(n)
-# 2	1	            1	    2
-# 3	1,2	            2	    1.5
-# 4	1,3	            2	    2
-# 5	1,2,3,4	            4	    1.25
-# 6	1,5	            2	    3
-# 7	1,2,3,4,5,6	    6	    1.1666...
-# 8	1,3,5,7	            4	    2
-# 9	1,2,4,5,7,8	    6	    1.5
+# n	    Relatively Prime    φ(n)    n/φ(n)
+# 2	    1	                1	    2
+# 3	    1,2	                2	    1.5
+# 4	    1,3	                2	    2
+# 5	    1,2,3,4	            4	    1.25
+# 6	    1,5	                2	    3
+# 7	    1,2,3,4,5,6	        6	    1.1666...
+# 8	    1,3,5,7	            4	    2
+# 9	    1,2,4,5,7,8	        6	    1.5
 # 10	1,3,7,9	            4	    2.5
 #
 # It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
diff --git a/Python/p081.py b/Python/p081.py
index da0289c..2cfab83 100644
--- a/Python/p081.py
+++ b/Python/p081.py
@@ -2,7 +2,7 @@
 
 # In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by only moving to the right and down,
 # is indicated in bold red and is equal to 2427.
-# 
+#
 #       *131* 673   234   103    18
 #       *201* *96* *342*  965   150
 #        630  803  *746* *422*  111
diff --git a/Python/p082.py b/Python/p082.py
index 560641a..cb3fe0e 100644
--- a/Python/p082.py
+++ b/Python/p082.py
@@ -11,7 +11,7 @@
 #       *201* *96* *342*  965   150
 #        630  803   746   422   111
 #        537  699   497   121   956
-#        805  732   524    37   331 
+#        805  732   524    37   331
 #
 # Find the minimal path sum, in matrix.txt, a 31K text file containing a 80 by 80 matrix, from the left column to the right column.
 
diff --git a/Python/p083.py b/Python/p083.py
index af4cbda..a689b5a 100644
--- a/Python/p083.py
+++ b/Python/p083.py
@@ -4,7 +4,7 @@
 #
 # In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by moving left, right, up, and down,
 # is indicated in bold red and is equal to 2297.
-# 
+#
 #       *131* 673  *234* *103*  *18*
 #       *201* *96* *342*  965  *150*
 #        630  803   746  *422* *111*
diff --git a/Python/p088.py b/Python/p088.py
index b43159c..0510a26 100644
--- a/Python/p088.py
+++ b/Python/p088.py
@@ -1,10 +1,12 @@
 #!/usr/bin/env python3
 
-# A natural number, N, hat can be written as the sum and product of a given set of at least two natural numbers, {a1,a2,...,ak} is called a product sum number: N = a1 + a2 + ... + ak = a1 x a2 x ... ak.
+# A natural number, N, hat can be written as the sum and product of a given set of at least two natural numbers, {a1,a2,...,ak} is called a product sum number:
+# N = a1 + a2 + ... + ak = a1 x a2 x ... ak.
 #
 # For example, 6 = 1 + 2 + 3 = 1 x 2 x 3
 #
-# For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k = 2,3,4,5, and 6 are as follows.
+# For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size,
+# k = 2,3,4,5, and 6 are as follows.
 #
 # k = 2: 4 = 2 x 2 = 2 + 2
 # k = 3: 6 = 1 x 2 x 3 = 1 + 2 + 3
diff --git a/Python/p090.py b/Python/p090.py
index d8b22ac..d4d1356 100644
--- a/Python/p090.py
+++ b/Python/p090.py
@@ -1,22 +1,26 @@
 #!/usr/bin/env python3
 
-# Each of the six faces on a cube has a different digit (0 to 9) written on it; the same is done to a second cube. By placing the two cubes side-by-side in different positions we can form a variety of 2-digit numbers.
+# Each of the six faces on a cube has a different digit (0 to 9) written on it; the same is done to a second cube. By placing the two cubes side-by-side
+# in different positions we can form a variety of 2-digit numbers.
 #
 # For example, the square number 64 could be formed:
 #                       |6||4|
 #
-# In fact, by carefully choosing the digits on both cubes it is possible to display all of the square numbers below one-hundred: 01, 04, 09, 16, 25, 36, 49, 64, and 81.
+# In fact, by carefully choosing the digits on both cubes it is possible to display all of the square numbers below one-hundred:
+# 01, 04, 09, 16, 25, 36, 49, 64, and 81.
 #
 # For example, one way this can be achieved is by placing {0,5,6,7,8,9} on one cube and {1,2,3,4,8,9} on the other cube.
 #
-# However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like  {0,5,6,7,8,9} and  {1,2,3,4,6,7} allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain 09.
+# However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like  {0,5,6,7,8,9} and  {1,2,3,4,6,7} allows
+# for all nine square numbers to be displayed; otherwise it would be impossible to obtain 09.
 #
 # In determining a distinct arrangement we are interested in the digits on each cube, not the order.
 #
 #   {1,2,3,4,5,6} is equivalent to {3,6,4,1,2,5}
 #   {1,2,3,4,5,6} is distinct from {1,2,3,4,5,9}
 #
-# But because we are allowing 6 and 9 to be reversed, the two distinct sets in the last example both represent the extended set {1,2,3,4,5,6,9} for the purpose of forming 2-digit numbers.
+# But because we are allowing 6 and 9 to be reversed, the two distinct sets in the last example both represent the extended set {1,2,3,4,5,6,9}
+# for the purpose of forming 2-digit numbers.
 #
 # How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?
 
diff --git a/Python/p091.py b/Python/p091.py
index 39a3faa..a0e414c 100644
--- a/Python/p091.py
+++ b/Python/p091.py
@@ -2,7 +2,8 @@
 
 # The points P(x1,y1) and Q(x2,y2) are plotted at integer co-ordinates and are joined to the origin, O(0,0), to form ΔOPQ.
 #
-# There are exactly fourteen triangles containing a right angle that can be formed when each co-ordinate lies between 0 and 2 inclusive; that is, 0<=x1,y1,x2,y2<=2.
+# There are exactly fourteen triangles containing a right angle that can be formed when each co-ordinate lies between 0 and 2 inclusive;
+# that is, 0<=x1,y1,x2,y2<=2.
 #
 # Given that 0<=x1,y1,x2,y2<=50, how many right triangles can be formed?
 
diff --git a/Python/p099.py b/Python/p099.py
index b55bf82..1833247 100644
--- a/Python/p099.py
+++ b/Python/p099.py
@@ -1,9 +1,9 @@
-# Comparing two numbers written in index form like 2^11 and 3^7 is not difficult, as any calculator would confirm that 
+# Comparing two numbers written in index form like 2^11 and 3^7 is not difficult, as any calculator would confirm that
 # 2^11 = 2048 < 3^7 = 2187.
 #
 # However, confirming that 632382^518061 > 519432^525806 would be much more difficult, as both numbers contain over three million digits.
 #
-# Using base_exp.txt, a 22K text file containing one thousand lines with a base/exponent pair on each line, 
+# Using base_exp.txt, a 22K text file containing one thousand lines with a base/exponent pair on each line,
 # determine which line number has the greatest numerical value.
 #
 # NOTE: The first two lines in the file represent the numbers in the example given above.*/
diff --git a/Python/projecteuler.py b/Python/projecteuler.py
index b704936..a86785c 100644
--- a/Python/projecteuler.py
+++ b/Python/projecteuler.py
@@ -294,7 +294,7 @@ def is_semiprime(n, primes):
         return False, -1, -1
 
     # Check if n is semiprime and one of the factors is 3.
-    elif n % 3 == 0:
+    if n % 3 == 0:
         if primes[n//3] == 1:
             p = 3
             q = n // 3