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Start your free trialEmmanuel Prempeh
1,769 PointsMy Codes for the Extra Credit Assignment
This is what I did for the extra credit. Is anyone with alternative approach? Cheers!
for (var i = 1; i <= 100; i++) {
//console.log(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)
}
};
6 Answers
Diego Villaseñor
12,615 PointsThis is mine, though it is not as elegant:
var counter = 0;
while (counter<100){
counter=counter+1;
if (counter%3) {
if (counter%5) {
console.log(counter);
} else {
console.log("Buzz");
}
} else {
if (counter%5) {
console.log("Fizz");
} else {
console.log("Fizzbuzz");
}
}
}
Kofi Owusu Acheampong
20,320 PointsThat looks pretty good. I did not do this particular task yet but I am guessing you are trying to find numbers divisible by 3 and 5 between 1 and 100 and print something to the screen if you find the number? If that is the case I guess you can also create a separate function? This should definitely be the primary way of going about it though.
Emmanuel Prempeh
1,769 Pointsthanks, Kofi
Mohammed Khalil Ait Brahim
9,539 PointsIf I may I thunk there is a small mistake there Let's say take 15 which is both divisible by 5 and 3 then Your if statements would output only "Fizz" Because (15 % 3 == 0) is true no ? And if you run this code I am sure no "fizzbuzz" would be printed to the screen. So the order of the if statement should change
if (i % 3 == 0 && i % 5 == 0) {
console.log("fizzbuzz")
}
else if (i % 5 == 0) {
console.log("buzz")
}
else if (i % 3 == 0) {
console.log("fizz")
}
else {
console.log(i)
}
Emmanuel Prempeh
1,769 PointsThank you for the correction!
Mohammed Khalil Ait Brahim
9,539 PointsAnytime !
Elizabeth Mackin
4,902 PointsHere is my code
for (var num = 100; num; num = num - 1) {
if (num % 15 == 0) {
console.log("fizzbuzz");
}
else if (num % 3 == 0) {
console.log("fizz");
}
else if(num % 5 == 0) {
console.log("buzz");
}
else {
console.log(num);
}
}