## and Answers"

FizzBuzz is a popular coding challenge that has been used in job interviews and programming classes for many years. It involves printing out numbers from 1 to 100, with certain conditions for multiples of 3 and 5. While it may seem like a simple task, there are actually many different variations and alternate questions that can be asked. In this article, we will explore some of these alternate FizzBuzz questions and provide answers to help you improve your coding skills.

## 1. FizzBuzz with a twist

The classic FizzBuzz question requires you to print "Fizz" for multiples of 3, "Buzz" for multiples of 5, and "FizzBuzz" for numbers that are both multiples of 3 and 5. But what if we change the multiples? For example, what if we print "Fizz" for multiples of 4 and "Buzz" for multiples of 7? The key to solving this is to use the modulus operator (%) to check for remainders. If a number has a remainder of 0 when divided by 4, then it is a multiple of 4 and we can print "Fizz". Similarly, if a number has a remainder of 0 when divided by 7, then it is a multiple of 7 and we can print "Buzz". If it is both, we can print "FizzBuzz". Here's the code:

## ```

for (int i = 1; i <= 100; i++) {

if (i % 4 == 0 && i % 7 == 0) {

## System.out.println("FizzBuzz");

## } else if (i % 4 == 0) {

## System.out.println("Fizz");

## } else if (i % 7 == 0) {

## System.out.println("Buzz");

## } else {

## System.out.println(i);

## }

## }

## ```

## 2. FizzBuzz without if statements

Another variation of FizzBuzz is to solve it without using if statements. This may seem like a daunting task, but it can actually be accomplished using the ternary operator. The ternary operator is a shorthand way of writing if/else statements. Here's how we can use it for FizzBuzz:

## ```

for (int i = 1; i <= 100; i++) {

String output = (i % 3 == 0) ? "Fizz" : "";

output += (i % 5 == 0) ? "Buzz" : "";

## System.out.println(output.isEmpty() ? i : output);

## }

## ```

In this code, we first check if the number is a multiple of 3. If it is, we assign "Fizz" to the output variable. Then, we use the += operator to append "Buzz" to the output if the number is also a multiple of 5. Finally, we check if the output is empty. If it is, we print the number. Otherwise, we print the output.

## 3. FizzBuzz with recursion

Recursion is a powerful concept in programming and can be used to solve FizzBuzz in a unique way. Instead of using a for loop, we can create a recursive method that prints out the numbers and calls itself until it reaches 100. Here's how it can be done:

## ```

## public static void fizzBuzz(int num) {

if (num % 3 == 0 && num % 5 == 0) {

## System.out.println("FizzBuzz");

## } else if (num % 3 == 0) {

## System.out.println("Fizz");

## } else if (num % 5 == 0) {

## System.out.println("Buzz");

## } else {

## System.out.println(num);

## }

## if (num < 100) {

## fizzBuzz(num + 1);

## }

## }

## ```

In this code, we first check for FizzBuzz, then Fizz, then Buzz, and finally print the number. Then, we call the method again with the next number until we reach 100.

## 4. FizzBuzz with a different range

So far, we have been printing numbers from 1 to 100. But what if we want to print a different range? For example, from 50 to 150? We can easily modify our code to do that:

## ```

for (int i = 50; i <= 150; i++) {

if (i % 3 == 0 && i % 5 == 0) {