Question

LANGUAGE C++
I need help with all 3 questions. Thank you in advance

Password Generator← 10 10.201.51. eRAD D scheduling山UitPro 6 Fizz Buzz Write a function "void FizzBuzz(int n)" that lists all of the numbers from 1 to n, but replace all numbers divisible by 3 but not divizible by 5 with the word "Fizz", replace all numbers divisible by 5 but not divisible by 3 with the word "Buzz", and replace all numbers divisible by both 3 and 5 with "FizzBuzz". The output for the command FizzBuzz(20)" should be as follows: Fizz Buzz Fizz Buzz Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz 7 Greatest Common Divisor Write a function "int god(int a, int b)" which uses the Euclidean algorithm to return the a and b. If you're not familiar with the Euclidean algorithm, google it, as you are required to use the Euclidean algorithm for this problem. You may assume that a 2 b>0 8 Prime numbers write the function tool i%Prim (int n" which returns true only if it is not divisible by any numher smaller than itself and greater than I. You may safely assume that n > I if n is prime, false otherwise. A number is prime if and DELL

Answer 1

Please find the code for 6, 7 and 8 below. Please read the comments carefully to understand the code.

6.

CODE

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#include <iostream>
using namespace std;

/*
* fizzBuzz() function takes an integer argument 'n' and performs the following operations
* 1. If a number is divisible by 15, print FizzBuzz
* 2. If a number is divisible by 3, print Fizz
* 3. If a number is divisible by 5, print Buzz
* 4. For all others, print the number
*/
void fizzBuzz(int n){
  
// Loop to traverse through all numbers between 1 and n
for(int i=1; i<=n; i++){
// if i is divisible by 15, print "FizzBuzz"
if(i % 15 == 0)
cout<<"FizzBuzz"<<endl;
// if i is divisible by 3, print "Fizz"
else if(i % 3 == 0)
cout<<"Fizz"<<endl;
// if i is divisible by 5, print "Buzz"
else if(i % 5 == 0)
cout<<"Buzz"<<endl;
// else, print the number
else
cout<<i<<endl;
}
}

int main() {
cout<<"Enter the value of n: ";
int n;
cin>>n;
//Call fizzBuzz() for n
fizzBuzz(n);
}

OUTPUT

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nter the value of n: 20 izz Buzz Fizz izz Buzz Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz ...Program finished with exit code 0 Press ENTER to exit console.0

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7.

CODE

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#include <iostream>
using namespace std;

/*
* The Euclidian GCD algorithm is given as.
* 1. If we subtract the smaller number(say b) from larger number(say a),
* GCD of a and b doesn’t change.
* 2. Hence, we keep subtracting repeatedly the smaller number from the larger number. Finally, we end up with GCD of a and b.
* 3. We can make the code easy if we use division instead of subtraction.
* 4. If we divide the larger number with the smaller number and get remainder 0, this means smaller number is the GCD.
*/

// GCD() function takes two integer arguments, a and b and fing their GCD using the above discussed logic.

int GCD (int a, int b) {
// if a is perfectly divisible by b, return b because GCD = b
if(a % b == 0)
return b;
// use the recusrsive approach
// Call GCD() recursively with b as the first argument and a % b as the second argument.
return GCD(b, a%b);
}

int main() {
cout<<"Enter the value of a and b: ";
int a, b;
cin>>a>>b;
//Call GCD() for a and b
cout<<"GCD of "<<a<<" and "<<b<<" is : "<<GCD(a, b);
}

OUTPUT

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Enter the value of a and b: 24 36 GCD of 24 and 36 is : 12 ...Program finished with exit code 0 Press ENTER to exit console.

8.

CODE

/////////////////////////////////////////////////////////////////

#include <iostream>
using namespace std;

/*
* A number is called prime if it is not divisible by any number other than 1 and itself. Some prime numbers are
* 2, 3, 5, 7, 11, 13, etc.
*/

// isPrime() function takes one integer argument n and returns true if it is prime, else false.

bool isPrime(int n) {
// Since, 2 is the only even prime number, we will first check if the number is 2.
// If n == 2, return true
if(n == 2)
return true;
// now, check if n is even. Clearly, if n is even, it is not prime as 2 is the only even prime number.
// If n is even, return false
if(n % 2 == 0)
return false;
  
// The following lines of code will be executed only when n is odd.
// Run a for loop and divide n with every odd number between 3 and n/2.
// If n is perfectly divisible by any of the odd number, then it is not prime, hence, return false.
for(int i=3; i<=n/2; i+=2) {
if(n % i == 0)
return false;
}
  
// The following line of code will be executed only when n is not 2 or even and it is not
// divisible by any odd number between 3 and n/2.
// This means that n is prime, hence, return true.
return true;
}

int main() {
cout<<"Enter the value of n: ";
int n;
cin>>n;
//Call isPrime() for n
if(isPrime(n) == true)
cout<<n<<" is a prime number.";
else
cout<<n<<" is not a prime number.";
}

OUTPUT

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Enter the value of n: 71 71 is a prime number. ...Program finished with exit code 0 Press ENTER to exit console.