Solution of problem 0004 added.

problem0004.js

@@ -0,0 +1,69 @@
+// problem004.js
+
+// A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
+
+// Find the largest palindrome made from the product of two 3-digit numbers.
+
+///////////////////////////////////////////////////////////////////////////////
+// NOTES ABOUT THE SOLUTION:
+// This solution cycles to test all pairs of factors between 111 and 999 that meet the condition of generating a palindrome and saves the largest found.
+// I think there must be another optimal solution to avoid testing all cases
+// cutting the loop around the largest factor pair
+// That's why I thought about doing the loop from highest to lowest.
+///////////////////////////////////////////////////////////////////////////////
+
+let bottom = 111
+let top = 999
+
+function isPalindrome(n) {
+    // console.log(`${ n.toString().split("").reverse().join("") }`);
+    // console.log(`${ n.toString()}`) ;
+
+    return n.toString().split("").reverse().join("") === n.toString();
+}
+
+// Test isPalindrome condition
+// console.log(isPalindrome('1000'));
+// console.log(isPalindrome('1001'));
+// console.log(isPalindrome('3443'));
+// console.log(isPalindrome('3435'));
+
+
+let i;
+let j;
+let foundi;
+let foundj;
+let foundPalindrome;
+
+// Find all cases
+i = top;
+do {
+
+    j = top;
+    do {
+
+        if( isPalindrome( j * i ) )
+        {   
+            console.log(`FOUND: ${i} x ${j} = ${j * i} is Palindrome`);
+
+            if(!foundPalindrome || (i*j) > foundPalindrome)
+            {
+                foundi = i;
+                foundj = j;
+                foundPalindrome = (i*j);
+            }
+
+        } else {
+            // console.log(`FOUND: ${i} x ${j} = ${j * i} is NOT Palindrome`);
+        }
+
+        j--;
+    }
+    while ( j >= bottom /*&& !(found1 && found2)*/)
+
+
+    i--;
+}
+while ( i >= bottom /* && !(found1 && found2) */ )
+
+console.log(`Largest Palindrome => ${foundi} 𝗑 ${foundj} = ${foundPalindrome}`);