FizzBuzz Challenge

When you go to get a job as a web developer, you may be given specific programming challenges. It's good to practice challenges like this not only for job interviews, but it sharpens your skills in general. You'll find that most challeneges have some kind of iteration involved. One of the most common challenges is the FizzBuzz challenge. This has to do with loops and conditionals, so I think that this is a good challenge at this point in the course.

Instructions:

• Print/log the numbers from 1 to 100
• For multiples of three print "Fizz" instead of the number
• For multiples of five print "Buzz"
• For numbers which are multiples of both three and five print "FizzBuzz".

Hints:

If you have taken the course up to this point, then you know how to write a loop and output/log something for each iteration. You also know how to check for a condition with "if/else/else if". You also know how to get a remainder of a number using the modulus operator (%). This is all you need to know to complete this challenge. Good luck!

Click For Solution

Solution 1: For Loop

  for (let i = 1; i <= 100; i++) {
    if (i % 15 === 0) {
    	console.log("FizzBuzz");
    } else if (i % 3 === 0) {
    	console.log("Fizz");
    } else if (i % 5 === 0) {
    	console.log("Buzz");
    } else {
    	console.log(i);
    }
}

In the above code, we set out initialize expression to 1. We set the condition to i <= 100. We set out increment expression to i++.

We first checked if i was divisible by 15. Because this means i is divisible by both 3 and 5. Since that is the case, we printed "FizzBuzz". Then we checked to see if i was divisible by 3. If so, we printed "Fizz". Then we checked to see if i was divisible by 5. If so, we printed "Buzz". If i was not divisible by either 3 or 5, we printed i (The current number).

Solution 2: While Loop

  let j = 1;

  while(j <= 100) {
    if (j % 15 === 0) {
    	console.log("FizzBuzz");
    } else if (j % 3 === 0) {
    	console.log("Fizz");
    } else if (j % 5 === 0) {
    	console.log("Buzz");
    } else {
    	console.log(j);
    }

    j++;
  }

We did the same thing here, just with a while loop