Question

C++

2. (a) Write a function, printdixisors, that takes a single integer parameter and prints all the numbers less that the parameter that are divisors of the parameter (i.e. divides it without a remainder) including 1. So printdivisors(6) will print 1,2,3. Note you may use a wrapper function or default parameters. (b) Write a function, sumdixisors, that takes a single integer parameter and returns the sum of all the divisors of the parameter (including 1). So sumdivisors(6) will return 6 as 1+2+3=6. Note you may use a wrapper function or default parameters. (c) Write a function, allperfects, that takes two parameters, each an integer, in any order and prints out all the perfect numbers between the lower parameter and the higher parameter. A perfect number is one is which the sum of its divisors is equal to the number itself. Remember to explain your functions, either by adding comments or showing how you derived the function. State any assumptions you make.

Answer 1

C++ program :-

#include <iostream>

using namespace std;

void printdivisors(int num){
for (int i=1;i<num;i++)
if (num%i==0)
cout<<i<<" ";
}

int sumdivisors(int num){
int sum=0;
for (int i=1;i<num;i++)
if (num%i==0)
sum=sum+i;

return sum;
}

void allperfects(int low_range, int high_range)
{
for(int i=low_range;i<=high_range;i++)
{
int sum=0;
for (int j = 1; j <= (i - 1); j++)
{
int rem = i % j;
   if (rem == 0)
{
sum = sum + j;
}
}
if (sum == i)
cout<<i<<" ";
}
}

int main()
{
cout<<"Print divisors :"<<endl;
printdivisors(6);
  
cout<<endl<<"The perfect numbers :"<<endl;
allperfects(1,20);
  
cout<<endl<<"The sum of divisors :"<<sumdivisors(6)<<endl;

}

Output :-

Ex Result CPU Time: 0.00 sec(s), Memory: 3324 kilobyte(s) Print divisors : 1 23 The perfect numbers : The sum of divisors :6