diff --git a/index.js b/index.js
index 6b0fec3ad..5f67a9dbb 100644
--- a/index.js
+++ b/index.js
@@ -1,7 +1,58 @@
 // Iteration 1: Names and Input
 
+// 1.1 Create a variable `hacker1` with the driver's name
+const hacker1 = "Joao";
 
-// Iteration 2: Conditionals
+// 1.2 Print "The driver's name is XXXX"
+console.log("The driver's name is " + hacker1);
+
+// 1.3 Create a variable `hacker2` with the navigator's name
+const hacker2 = "JC";
 
+// 1.4 Print "The navigator's name is YYYY"
+console.log(`The navigator's name is ${hacker2}`);
+
+// Iteration 2: Conditionals
 
+// 2.1. Depending on which name is longer, print:
+if (hacker1.length > hacker2.length) {
+  console.log(
+    `The driver has the longest name, it has ${hacker1.length} characters.`,
+  );
+} else if (hacker1.length < hacker2.length) {
+  console.log(
+    `It seems that the navigator has the longest name, it has ${hacker2.length} characters.`,
+  );
+} else {
+  console.log(
+    `Wow, you both have equally long names, ${hacker1.length} characters!`,
+  );
+}
 // Iteration 3: Loops
+
+// 3.1 Print the characters of the driver's name, separated by space, and in capital letters, i.e., "J O H N".
+let nameCapital = "";
+for (let i = 0; i < hacker1.length; i++) {
+  if (i < hacker1.length - 1) {
+    nameCapital += hacker1[i].toUpperCase() + " ";
+  } else {
+    nameCapital += hacker1[i].toUpperCase();
+  }
+}
+console.log(nameCapital);
+
+// 3.2 Print all the characters of the navigator's name in reverse order, i.e., "nhoJ".
+let nameReverse = "";
+for (let i = hacker1.length - 1; i >= 0; i--) {
+  nameReverse += hacker1[i];
+}
+console.log(nameReverse);
+
+// 3.3 Depending on the lexicographic order of the strings, print:
+if (hacker1 < hacker2) {
+    console.log("The driver's name goes first.");
+    } else if (hacker1 > hacker2) {
+        console.log("Yo, the navigator goes first definitely.");
+    } else {
+        console.log("What?! You both have the same name?");
+    }