From abcb2bc3888ffc91dbab3a12c5a76f019439191d Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Daniel=20L=C3=B8vbr=C3=B8tte=20Olsen?=
 <daniel.olsen99+GitHub@gmail.com>
Date: Wed, 30 Dec 2015 15:28:28 +0100
Subject: [PATCH] Added mathermatical solution

---
 Project-Euler/001/ArithmeticProgression.c | 36 +++++++++++++++++++++++
 1 file changed, 36 insertions(+)
 create mode 100644 Project-Euler/001/ArithmeticProgression.c

diff --git a/Project-Euler/001/ArithmeticProgression.c b/Project-Euler/001/ArithmeticProgression.c
new file mode 100644
index 0000000..5600bf3
--- /dev/null
+++ b/Project-Euler/001/ArithmeticProgression.c
@@ -0,0 +1,36 @@
+#include <stdio.h>
+#include <string.h>
+#include <math.h>
+#include <stdlib.h>
+
+// Use:
+// bruteforce
+// n numbers to calculate (PE 1)
+// number #1 to calculate (PE 1000)
+// number #2 to calculate
+// ...
+// number n to calculate
+
+// Prototype
+long sum_of_divisible(long divisible, long last);
+
+int main() {
+    int n; 
+    scanf("%d",&n);
+    long arr[n];
+    for(int i = 0; i < n; i++){
+       scanf("%li",&arr[i]);
+    }
+    
+    for (int i = 0; i < n; i++)
+    {
+        printf("%li\n", sum_of_divisible(3, arr[i] - 1) + sum_of_divisible(5, arr[i] - 1) - sum_of_divisible(15, arr[i] - 1));
+    }
+}
+
+
+// finds the sum of all the numbers divisible by "divisible" up to and including "last" using Arithmetic Progression
+long sum_of_divisible(long divisible, long last)
+{
+    return (long) divisible*(last/divisible)*((last/divisible)+1)/2;
+}