completes aplify

Author: umbur

amplify.js

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+// In this challenge, you must generate a sequence of consecutive numbers, from a lower bound that will always be equal to 1, up to
+// a variable given higher bound (including the bounds in the sequence).
+// Each number of the sequence that can be exactly divided by 4 must be amplified by 10 (see notes below).
+// Given a higher bound num, implement a function that returns an array with the sequence of numbers, after that every multiple of
+// 4 has been amplified.
+
+// Examples:
+// amplify(4) ➞ [1, 2, 3, 40]
+// Create a sequence from 1 to 4
+// 4 is exactly divisible by 4, so it will be 4*10 = 40
+// amplify(3) ➞ [1, 2, 3]
+// Create a sequence from 1 to 3
+// There are no numbers that can be exactly divided by 4
+// amplify(25) ➞ [1, 2, 3, 40, 5, 6, 7, 80, 9, 10, 11, 120, 13, 14, 15, 160, 17, 18, 19, 200, 21, 22, 23, 240, 25]
+// Create a sequence from 1 to 25
+// The numbers exactly divisible by 4 are: 4 (4*10 = 40), 8 (8 * 10 = 80)... and so on.
+
+// Notes:
+// The given parameter num will always be equal to or greater than 1.
+// Remember to include the num as the higher bound of the sequence (see the Examples) above.
+// A number a amplified by a factor b can also be read as: a * b.
+// A number a is exactly divisible by a number b when the remainder of the division a / b is equal to 0.
+// If you get stuck on a challenge, find help in the Resources tab.
+// If you're really stuck, unlock solutions in the Solutions tab.
+
+function amplify(num) {
+  let result = [];
+  for (let i = 1; i <= num; i++) {
+    if (i % 4 == 0) {
+      result.push(i * 10);
+    } else {
+      result.push(i);
+    }
+  }
+  return result;
+}
+console.log(amplify(25));

chatroomStatus.js

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+// Write a function that returns the number of users in a chatroom based on the following rules:
+// If there is no one, return "no one online".
+// If there is 1 person, return "user1 online".
+// If there are 2 people, return "user1 and user2 online".
+// If there are n>2 people, return the first two names and add "and n-2 more online".
+// For example, if there are 5 users, return:
+// "user1, user2 and 3 more online"
+
+// Examples:
+// chatroomStatus([]) ➞ "no one online"
+// chatroomStatus(["paRIE_to"]) ➞ "paRIE_to online"
+// chatroomStatus(["s234f", "mailbox2"]) ➞ "s234f and mailbox2 online"
+// chatroomStatus(["pap_ier44", "townieBOY", "panda321", "motor_bike5", "sandwichmaker833", "violinist91"])
+// ➞ "pap_ier44, townieBOY and 4 more online"
+
+function chatroomStatus(users) {
+  //   console.log(users.length);
+  if (users.length === 0) {
+    return "no one online";
+  } else if (users.length === 1) {
+    return `${users[0]} online`;
+  } else if (users.length === 2) {
+    return `${users[0]} and ${users[1]} online`;
+  } else {
+    return `${users[0]}, ${users[1]} and ${users.length - 2} more online`;
+  }
+}
+console.log(
+  chatroomStatus([
+    "pap_ier44",
+    "townieBOY",
+    "panda321",
+    "motor_bike5",
+    "sandwichmaker833",
+    "violinist91",
+  ])
+);