Java Program to Check Perfect Number with Output

A Java program to check a perfect number sums all its positive divisors except itself using a loop and compares the sum to the number with code like for (int i = 1; i <= num / 2; i++) { if (num % i == 0) sum += i; } and then checks if (sum == num).

Examples

Input6

Output6 is a perfect number

Input28

Output28 is a perfect number

Input10

Output10 is not a perfect number

How to Think About It

To check if a number is perfect, find all numbers smaller than it that divide it evenly. Add these divisors together. If the total equals the original number, it is perfect; otherwise, it is not.

Algorithm

1

Get the input number.

2

Initialize a sum to zero.

3

Loop from 1 to half of the number.

4

If the current number divides the input evenly, add it to the sum.

5

After the loop, compare the sum with the input number.

6

If they are equal, the number is perfect; otherwise, it is not.

Code

java

import java.util.Scanner;

public class PerfectNumber {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int num = sc.nextInt();
        int sum = 0;
        for (int i = 1; i <= num / 2; i++) {
            if (num % i == 0) sum += i;
        }
        if (sum == num) System.out.println(num + " is a perfect number");
        else System.out.println(num + " is not a perfect number");
        sc.close();
    }
}

Output

6 is a perfect number

Dry Run

Let's trace input 6 through the code

1

Input number

num = 6, sum = 0

2

Loop from 1 to 3 (6/2)

Check divisors 1, 2, 3

3

Check divisor 1

6 % 1 == 0, sum = 0 + 1 = 1

4

Check divisor 2

6 % 2 == 0, sum = 1 + 2 = 3

5

Check divisor 3

6 % 3 == 0, sum = 3 + 3 = 6

6

Compare sum and number

sum = 6 equals num = 6, so print perfect

Divisor	Is Divisor?	Sum
1	Yes	1
2	Yes	3
3	Yes	6

Why This Works

Step 1: Find divisors

We check all numbers less than or equal to half the input because no divisor (except the number itself) can be greater than half.

Step 2: Sum divisors

We add each divisor to a sum to find the total of all proper divisors.

Step 3: Compare sum to number

If the sum equals the original number, it means the number is perfect by definition.

Alternative Approaches

Using while loop

java

import java.util.Scanner;

public class PerfectNumber {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int num = sc.nextInt();
        int sum = 0, i = 1;
        while (i <= num / 2) {
            if (num % i == 0) sum += i;
            i++;
        }
        if (sum == num) System.out.println(num + " is a perfect number");
        else System.out.println(num + " is not a perfect number");
        sc.close();
    }
}

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

Using method to check perfect number

java

import java.util.Scanner;

public class PerfectNumber {
    public static boolean isPerfect(int num) {
        int sum = 0;
        for (int i = 1; i <= num / 2; i++) {
            if (num % i == 0) sum += i;
        }
        return sum == num;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int num = sc.nextInt();
        if (isPerfect(num)) System.out.println(num + " is a perfect number");
        else System.out.println(num + " is not a perfect number");
        sc.close();
    }
}

Separates logic into a method for better code reuse and clarity.

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

Time Complexity

The program loops from 1 to n/2 to find divisors, so it runs in linear time O(n).

Space Complexity

Uses only a few variables for sum and loop counters, so space is constant O(1).

Which Approach is Fastest?

All approaches check divisors similarly; using a method improves readability but not speed.

Approach	Time	Space	Best For
For loop	O(n)	O(1)	Simple and clear code
While loop	O(n)	O(1)	Shows loop variety
Method extraction	O(n)	O(1)	Code reuse and clarity