Create Solution_0001.js

Solution_0001.js

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+/* 1.
+ *
+ * Very simple brute force solution that goes
+ * through all of the numbers between 3..n and
+ * adds them up when they're divisible by 3 and 5.
+ */
+
+let n = 1000;
+let bruteforce_sum = 0;
+
+// Start at i = 3, since we're not interested in anything below 3
+for(let i = 3; i < n; i++) {
+
+  bruteforce_sum += i % 3 === 0 || i % 5 === 0 ? i : 0;
+
+}
+
+console.log("Bruteforce Solution:", bruteforce_sum); // 233168
+
+
+
+/* 2.
+ *
+ * Way more efficient solution using finite arithmetic progression and series
+ */
+
+function sum(n) {
+  return (n * (n + 1)) / 2;
+}
+
+const threes = (3 * sum(Math.floor(999/3)));
+const fives = (5 * sum(Math.floor(999/5)));
+const fifteens = (15 * sum(Math.floor(999/15)));
+
+const solution =  threes + fives - fifteens;
+
+console.log("Solution:", solution); // 233168