C Program to Find Factors of a Number

To find factors of a number in C, use a for loop from 1 to the number and print all values that divide the number evenly using if (num % i == 0).

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 dividing the number by it leaves no remainder. If yes, 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 (remainder zero).

4

If yes, print that number as a factor.

5

End the loop after reaching the input number.

Code

c

#include <stdio.h>

int main() {
    int num, i;
    printf("Enter a number: ");
    scanf("%d", &num);
    printf("Factors of %d are: ", num);
    for(i = 1; i <= num; i++) {
        if(num % i == 0) {
            printf("%d ", i);
        }
    }
    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

num = 6

2

Start loop from 1 to 6

i = 1 to 6

3

Check if 6 % i == 0

For i=1: 6 % 1 = 0 (print 1) For i=2: 6 % 2 = 0 (print 2) For i=3: 6 % 3 = 0 (print 3) For i=4: 6 % 4 = 2 (skip) For i=5: 6 % 5 = 1 (skip) For i=6: 6 % 6 = 0 (print 6)

i	num % i	Print factor?
1	0	Yes (1)
2	0	Yes (2)
3	0	Yes (3)
4	2	No
5	1	No
6	0	Yes (6)

Why This Works

Step 1: Loop through numbers

The program checks every number from 1 to the input number to find possible factors.

Step 2: Check divisibility

Using num % i == 0 tests if the number divides evenly with no remainder.

Step 3: Print factors

If the condition is true, the current number is a factor and gets printed.

Alternative Approaches

Check factors up to half the number

c

#include <stdio.h>

int main() {
    int num, i;
    printf("Enter a number: ");
    scanf("%d", &num);
    printf("Factors of %d are: 1 ", num);
    for(i = 2; i <= num / 2; i++) {
        if(num % i == 0) {
            printf("%d ", i);
        }
    }
    if(num != 1) printf("%d", num);
    return 0;
}

This reduces the loop range to half the number since no factor (except the number itself) can be greater than half, improving efficiency.

Check factors using square root optimization

c

#include <stdio.h>
#include <math.h>

int main() {
    int num, i;
    printf("Enter a number: ");
    scanf("%d", &num);
    printf("Factors of %d are: ", num);
    int limit = (int)sqrt(num);
    for(i = 1; i <= limit; i++) {
        if(num % i == 0) {
            printf("%d ", i);
            if(i != num / i) printf("%d ", num / i);
        }
    }
    return 0;
}

This method checks up to the square root and prints both factors at once, making it faster for large numbers but factors may print unordered.

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

Time Complexity

The program loops from 1 to the input number, so it runs in linear time O(n). Each iteration does a simple modulus check.

Space Complexity

The program uses a fixed amount of memory for variables and prints results directly, so space complexity is O(1).

Which Approach is Fastest?

Using the square root optimization reduces time to about O(√n), which is faster for large numbers compared to the simple O(n) loop.

Approach	Time	Space	Best For
Simple loop 1 to n	O(n)	O(1)	Small to medium numbers, easy to understand
Loop 1 to n/2	O(n/2) ~ O(n)	O(1)	Slightly faster, still simple
Square root optimization	O(√n)	O(1)	Large numbers, better performance