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JavaScript

Posted by Jordan Hammond
Jordan Hammond
2,278 Points

Write a program that loops through the numbers 1 through 100. Each number should be printed to the console.

Can someone review my code and tell me why it won't work. I know how to get the right answer after looking at other posts but I still don't understand why this won't work.

It's from this extra credit question: Write a program that loops through the numbers 1 through 100. Each number should be printed to the console, using console.log(). However, if the number is a multiple of 3, don't print the number, instead print the word "fizz". If the number is a multiple of 5, print "buzz" instead of the number. If it is a multiple of 3 and a multiple of 5, print "fizzbuzz" instead of the number.

Thanks!

var counter=100

while(counter) {
    if(counter%15=0){
        console.log("fizzbuzz")
        counter=counter-1
    } else {
        if(counter%3=0){
            console.log("fizz")
            counter=counter-1
        } else {
            if(counter%5=0) {
                console.log("buzz")
                counter=counter-1
            } else {
                console.log("number " +counter)
                counter=counter-1
            }
        }
    }
}
Jordan Hammond
Jordan Hammond
2,278 Points

Nevermind. Figured it out.

my

counter%3=0

needs an extra =

so it should look like this:

counter%3==0

3 Answers

Ben Stoltz
Ben Stoltz
20,364 Points
Ben Stoltz
Ben Stoltz
20,364 Points

All of your modulo (%) statements are only using one =. == will work but it might be a good habit to get into with using ===.

== : Will compare both values, but will convert type. So '2' == 2 will return True.

===: Strict comparison '2' === 2 will return False.

It's just a good habit to get into.

Bassem Al Hassouni
PLUS
Bassem Al Hassouni
Courses Plus Student 3,395 Points
Bassem Al Hassouni
Bassem Al Hassouni
Courses Plus Student 3,395 Points

can it be done like this

for (var i = 0; i < 100; i += 1) {

if (i % 15 === 0) {

    console.log("fizzbuzz");

} else if (i % 3 === 0) {

    console.log("fizz");

} else if (i % 5 === 0) {

    console.log("buzz");

} else if (i % 3 > 0 || i % 5 > 0) {

    console.log(i);
}

}

Marisa Hendrickson
Marisa Hendrickson
8,303 Points
Marisa Hendrickson
Marisa Hendrickson
8,303 Points

You can also do it this way

for (var counter = 100;counter;counter = counter - 1) {

if (counter % 15 === 0) {
    console.log("fizzbuzz");
} else if (counter % 3 === 0) {
    console.log("Fizz");
} else if(counter % 5 === 0){
    console.log("buzz");   
} else {
    console.log(counter);
}

}