Java Program to Find Factors of a Number

To find factors of a number in Java, use a for loop from 1 to the number and check if the number is divisible by the loop variable using if (number % i == 0) to print the factors.

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 with 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

java

import java.util.Scanner;

public class Factors {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int number = sc.nextInt();
        System.out.print("Factors of " + number + " are: ");
        for (int i = 1; i <= number; i++) {
            if (number % i == 0) {
                System.out.print(i + " ");
            }
        }
        sc.close();
    }
}

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 number = 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

5

End loop

Loop ends after i = 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 use a for loop to check every number from 1 to the input number because factors must be within this range.

Step 2: Check divisibility

Using number % i == 0 checks if the remainder is zero, meaning i divides the number exactly.

Step 3: Print factors

When the condition is true, we print i as a factor because it divides the number evenly.

Alternative Approaches

Using while loop

java

import java.util.Scanner;

public class FactorsWhile {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int number = sc.nextInt();
        int i = 1;
        System.out.print("Factors of " + number + " are: ");
        while (i <= number) {
            if (number % i == 0) {
                System.out.print(i + " ");
            }
            i++;
        }
        sc.close();
    }
}

Uses a while loop instead of for loop; functionally same but shows different loop style.

Optimized up to square root

java

import java.util.Scanner;

public class FactorsOptimized {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int number = sc.nextInt();
        System.out.print("Factors of " + number + " are: ");
        for (int i = 1; i <= Math.sqrt(number); i++) {
            if (number % i == 0) {
                System.out.print(i + " ");
                if (i != number / i) {
                    System.out.print((number / i) + " ");
                }
            }
        }
        sc.close();
    }
}

Checks factors only up to square root for better performance, printing both factors at once.

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

Time Complexity

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

Space Complexity

It uses only a few variables and prints output directly, so space is constant O(1).

Which Approach is Fastest?

The optimized approach checking up to square root of n reduces time to O(√n), which is faster for large numbers.

Approach	Time	Space	Best For
Simple for loop	O(n)	O(1)	Small to medium numbers
While loop	O(n)	O(1)	Same as for loop, different style
Optimized sqrt loop	O(√n)	O(1)	Large numbers for better performance