[ADD] smallest common added

luisprooc · luisprooc · commit c6c76a4c0355 · 2021-07-14T12:00:35.000-04:00
diff --git a/README.md b/README.md
@@ -699,4 +699,29 @@ dropElements([1, 2, 3, 7, 4], function(n) {return n > 3;}) should return [7, 4].
 
 dropElements([1, 2, 3, 9, 2], function(n) {return n > 2;}) should return [3, 9, 2].
 
+```
+
+
+## Smallest Common Multiple
+
+Find the smallest common multiple of the provided parameters that can be evenly divided by both, as well as by all sequential numbers in the range between these parameters.
+
+The range will be an array of two numbers that will not necessarily be in numerical order.
+
+For example, if given 1 and 3, find the smallest common multiple of both 1 and 3 that is also evenly divisible by all numbers between 1 and 3. The answer here would be 6.
+
+
+```javascript
+smallestCommons([1, 5]) should return a number.
+
+smallestCommons([1, 5]) should return 60.
+
+smallestCommons([5, 1]) should return 60.
+
+smallestCommons([2, 10]) should return 2520.
+
+smallestCommons([1, 13]) should return 360360.
+
+smallestCommons([23, 18]) should return 6056820.
+
 ```
diff --git a/src/smallestCommon.js b/src/smallestCommon.js
@@ -0,0 +1,21 @@
+function commonDivisor(arr,divisor){
+    for(let [small,big] = arr;small <= big; ++small){
+        if(divisor % small !== 0) return false;
+    }
+    return true;
+}
+
+function smallestCommons(arr) {
+    arr = arr.sort((a,b) => a-b);
+    let [small,big] = arr;
+    let divisor = big * 2;
+    while(true){
+        if(divisor % small === 0 && divisor % big === 0){
+            if(commonDivisor(arr,divisor)) return divisor;
+        }
+        ++divisor;
+    }
+
+}
+  
+console.log(smallestCommons([2,10]));
