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Added comments to the python code for the first 25 problems

C/p003.c

@@ -4,7 +4,6 @@
 
 #include <stdio.h>
 #include <stdlib.h>
-#include <math.h>
 #include <time.h>
 #include "projecteuler.h"
 

C/p011.c

@@ -1,4 +1,4 @@
-/* n the 20×20 grid below, four numbers along a diagonal line have been marked in red.
+/* In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
  *
  * 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
  * 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00

C/p020.c

@@ -33,7 +33,7 @@ int main(int argc, char **argv)
 
    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 20\n");
    gmp_printf("Answer: %Zd\n", sum);

C/p021.c

@@ -3,8 +3,8 @@
  *
  * For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284.
  * The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
-
-Evaluate the sum of all the amicable numbers under 10000.*/
+ *
+ * Evaluate the sum of all the amicable numbers under 10000.*/
 
 #include <stdio.h>
 #include <stdlib.h>