Program to find the divisors of a positive Integer

Problem Statement

We need to write a program that accepts a positive integer value from the user and prints all the possible divisors of that number (excluding the number itself).

For example

Input: 100
Output : 1 2 4 5 10 25 50

Algorithm

The problem statement is quite straightforward and beginner-level. To solve this problem, all we need is a loop and a conditional if statement. As the problem statement state that all we need to do is find all the divisors except for the number itself.

So we need a loop from range 1 to half of the number entered by the user. Because a number's all possible divisors can be the number itself and less than half of it.

C Program to find the divisors of a Positive Integer

#include <stdio.h>

int main()
{	
	int num, i;
	//input the number
	printf("Enter a positive integer value: ");
	scanf("%d",&num);
	
	for(i=1; i<=num/2; i++)
	{
		if(num%i==0)
		{
			printf("%d ", i);
		}
	}
	

	return 0;
}

Output

Enter a positive integer value: 100
1 2 4 5 10 20 25 50

C++ program to find the divisors of a Positive Integer

#include<iostream>

using namespace std;
int main()
{
	int num;
	
	//input the positive integer
	cout<<"Enter a Positive Integer Number: "; cin>>num;
	
	for(int i=1; i<=num/2; i++)
	{
		if(num%i==0)
		{
			cout<<i<<" ";
		}
	}
	return 0; 
}

Output:

Enter a Positive Integer Number: 100
1 2 4 5 10 20 25 50

Python:

# input number
num = int(input("Enter a Positive Integer Number: "))

for i in range(1, (num//2)+1):
    if num%i==0:
        print(i, end =" ")

Output:

Enter a Positive Integer Number: 100
1 2 4 5 10 20 25 50

Wrapping Up!

Now let's wrap up this programming tutorial on "Write a program to find the divisors of a positive integer". To find all the divisors of a number n we just need to calculate all smaller numbers that can divide the number n completely.

If we look closely, all the divisor of a number lies between 1 to half of the number itself, which save time, and we just need to create a loop that would iterate n/2 time.

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