C++ Program to Find Factors of a Number

You can find factors of a number in C++ by looping from 1 to the number and checking if number % i == 0, then printing i as a factor.

Examples

Input6

OutputFactors of 6 are: 1 2 3 6

Input13

OutputFactors of 13 are: 1 13

Input1

OutputFactors of 1 are: 1

How to Think About It

To find factors of a number, think about all numbers from 1 up to that number. For each, check if it divides the number evenly using the remainder operator %. If the remainder is zero, that number is a factor.

Algorithm

1

Get the input number from the user.

2

Start a loop from 1 to the input number.

3

For each number in the loop, check if it divides the input number evenly using <code>number % i == 0</code>.

4

If yes, print that number as a factor.

5

End the loop after reaching the input number.

Code

cpp

#include <iostream>
using namespace std;

int main() {
    int number;
    cout << "Enter a number: ";
    cin >> number;
    cout << "Factors of " << number << " are: ";
    for (int i = 1; i <= number; ++i) {
        if (number % i == 0) {
            cout << i << " ";
        }
    }
    cout << endl;
    return 0;
}

Output

Enter a number: 6
Factors of 6 are: 1 2 3 6 

Dry Run

Let's trace the input 6 through the code to find its factors.

1

Input number

User enters 6.

2

Start loop

Loop variable i starts at 1.

3

Check divisibility

Check if 6 % i == 0 for i = 1 to 6.

4

Print factors

Print i when condition is true: 1, 2, 3, 6.

i	6 % i	Is factor?
1	0	Yes
2	0	Yes
3	0	Yes
4	2	No
5	1	No
6	0	Yes

Why This Works

Step 1: Loop through numbers

We check every number from 1 to the input number to see if it divides the number evenly.

Step 2: Use modulo operator

The % operator gives the remainder. If remainder is zero, the number divides evenly.

Step 3: Print factors

All numbers that divide evenly are printed as factors.

Alternative Approaches

Check factors up to square root

cpp

#include <iostream>
#include <cmath>
using namespace std;

int main() {
    int number;
    cout << "Enter a number: ";
    cin >> number;
    cout << "Factors of " << number << " are: ";
    for (int i = 1; i <= sqrt(number); ++i) {
        if (number % i == 0) {
            cout << i << " ";
            if (i != number / i) {
                cout << number / i << " ";
            }
        }
    }
    cout << endl;
    return 0;
}

This method is faster for large numbers because it only checks up to the square root, printing both factors at once.

Store factors in a vector and sort

cpp

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int number;
    vector<int> factors;
    cout << "Enter a number: ";
    cin >> number;
    for (int i = 1; i <= number; ++i) {
        if (number % i == 0) {
            factors.push_back(i);
        }
    }
    sort(factors.begin(), factors.end());
    cout << "Factors of " << number << " are: ";
    for (int f : factors) {
        cout << f << " ";
    }
    cout << endl;
    return 0;
}

This approach collects factors first, then prints them sorted. Useful if you want to process factors later.

Complexity: O(n) time, O(1) space

Time Complexity

The program checks every number from 1 to n, so it runs in O(n) time.

Space Complexity

It uses constant extra space, O(1), as it prints factors directly without storing them.

Which Approach is Fastest?

Checking up to the square root of n reduces time to O(√n), which is faster for large numbers.

Approach	Time	Space	Best For
Simple loop 1 to n	O(n)	O(1)	Small to medium numbers
Loop up to sqrt(n)	O(√n)	O(1)	Large numbers, faster execution
Store and sort factors	O(n log n)	O(n)	When factors need further processing