September 2017 Update

Task/Sum-multiples-of-3-and-5/JavaScript/sum-multiples-of-3-and-5-7.js

@@ -1,35 +1,36 @@
 (() => {
 
     // Area under straight line
-    // between first multiple and last
-    let sumMults = (n, factor) => {
-            let n1 = Math.floor((n - 1) / factor);
+    // between first multiple and last.
 
-            return Math.floor(factor * n1 * (n1 + 1) / 2);
-        },
+    // sumMults :: Int -> Int -> Int
+    const sumMults = (n, factor) => {
+        const n1 = quot(n - 1, factor);
+        return quot(factor * n1 * (n1 + 1), 2);
+    };
 
-        sum35 = (n) => sumMults(n, 3) + sumMults(n, 5) - sumMults(n, 15);
+    // sum35 :: Int -> Int
+    const sum35 = n => sumMults(n, 3) + sumMults(n, 5) - sumMults(n, 15);
 
 
-    // TEST
+    // GENERIC ----------------------------------------------------------------
 
-    // range(intFrom, intTo, optional intStep)
-    // Int -> Int -> Maybe Int -> [Int]
-    let range = (m, n, step) => {
-        let d = (step || 1) * (n >= m ? 1 : -1);
+    // enumFromTo :: Int -> Int -> [Int]
+    const enumFromTo = (m, n) =>
+        Array.from({
+            length: Math.floor(n - m) + 1
+        }, (_, i) => m + i);
 
-        return Array.from({
-            length: Math.floor((n - m) / d) + 1
-        }, (_, i) => m + (i * d));
-    }
+    // Integral a => a -> a -> a
+    const quot = (n, m) => Math.floor(n / m);
+
+    // TEST -------------------------------------------------------------------
 
     // Sums for 10^1 thru 10^8
-
-    return range(1, 8)
+    return enumFromTo(1, 8)
         .map(n => Math.pow(10, n))
         .reduce((a, x) => (
             a[x.toString()] = sum35(x),
             a
         ), {});
-
 })();