# FIzzBuzz exercise 2

#1

so this code returned stuff but I still got an error too!! any ideas what that error is referring to.

Oops, try again. It looks like you printed out the wrong number of items.

``````for(var i = 0; i <= 20; i ++) {

if ((i % 3 === 0) || (i % 5 === 0)) {
console.log("Fizz") || ("Buzz");

}
else {
console.log("FizzBuzz");

}
};

I also tried this version:
for(var i = 0; i < 21; i ++) {

if (i % 3 === 0) {
console.log("Fizz");
}
else if (i % 5 === 0) {
console.log("Buzz");
}
else {
console.log("FizzBuzz");

}
};``````

#2

This is the correct output:

``````1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
Fizz
19
Buzz``````

You have to implement this part of instructions:

Otherwise, just print out the number.

By otherwise it means that if number is not divisible by 3 and is not divisible by 5.

#3

My input is as follows:
Fizz
1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
Fizz
16
17
Fizz
19
Buzz

I don't see any FizzBuzz!! not sure why that didn't return any value?

This is my code:
for(var i = 0; i < 21; i ++) {

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("FizzBuzz");
}
else {
console.log(i);

}
};

any suggestions?

#4

Ok, let's say that we are inside the `for` loop and `i = 15`.

Condition of the `if` is `i % 3 === 0` -> `15 % 3 === 0` -> `0 === 0` -> `true`. So interpreter will print out `"Fizz"` and will leave the `if ... else if ... else` construction.

You have to rearrange your code. You should check if a number is divisible by `3` and by `5` in the first place.

#6

you want to put the if its divisible by 3 and 5 to the top of the other IFs. Otherwise, the 2 IFs that console fizz and buzz must give more specific argument such as:
if(i % 3 === 0 && i % 5 !==0) {
console.log("Fizz")
}
if "i" is divisible by 5 and "i" is not divisible by three, console log Buzz

#7

#9

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