Implement "Problem 6"

main.c

@@ -168,13 +168,36 @@ void problem_5() {
 /**
  * Problem 6
  *
- *
+ * Declare a float of "1e20".
+ * Print to console.
+ * Declare a float of "(1e20 + 3500000000)".
+ * Print to console.
+ * Declare a float of "(1e20 + (3500000000 * 1000000000))".
+ * Print to console.
+ * Declare a float of "1e20".
+ * Using a loop, add 3500000000 to the float, 1000000000 different times.
+ * Print to console.
  */
 void problem_6() {
     printf("\n\nProblem 6 start.\n\n");
     printf("Expected Output:\n");
-    printf("\n");
+    printf("100000002004087734272\n");
+    printf("100000002007587734272\n");
+    printf("103500002052240572416\n");
+    printf("103500002052240572416\n");
     printf("Actual Output:\n");
 
+    float test_float = 1e20;
+    printf("%.10lf\n", test_float);
+    test_float = (1e20 + 3500000000);
+    printf("%.10lf\n", test_float);
+    test_float = (1e20 + (3500000000 * 1000000000));
+    printf("%.10lf\n", test_float);
+    test_float = 1e20;
+    for (int index = 0; index < 1000000000; index++) {
+        test_float += 3500000000;
+    }
+    printf("%.10lf\n", test_float);
+
     printf("\nProblem 6 end.\n");
 }