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Perfect Number

Java coding interview problem for Number Logic: Perfect Number.

Checking whether a number is a Perfect Number is one of the classic Java coding interview questions.

This problem helps interviewers evaluate your understanding of:

  • Loops
  • Conditional statements
  • Divisibility
  • Number theory
  • Mathematical logic
  • Problem-solving skills

Perfect Number problems are common in coding interviews, competitive programming, and mathematics.

What is a Perfect Number?

A Perfect Number is a positive integer that is equal to the sum of all its positive divisors excluding itself.

In other words,

Perfect Number

=

Sum of Proper Divisors

A proper divisor is any positive divisor of a number except the number itself.

Mathematical Definition

If

N

has proper divisors

d1, d2, d3 ...

then

d1 + d2 + d3 + ...

=

N

Example 1

Number

6

Divisors

1

2

3

Sum

1 + 2 + 3

=

6

Since

Sum = Number

Therefore

6 is a Perfect Number

Example 2

Number

28

Divisors

1

2

4

7

14

Sum

1 + 2 + 4 + 7 + 14

=

28

Therefore

28 is a Perfect Number

Example 3

Number

12

Divisors

1

2

3

4

6

Sum

1 + 2 + 3 + 4 + 6

=

16

Since

16 ≠ 12

Therefore

12 is NOT a Perfect Number

Some Perfect Numbers

Number Perfect?
6 ✅ Yes
28 ✅ Yes
496 ✅ Yes
8128 ✅ Yes
12 ❌ No
20 ❌ No
100 ❌ No

Real Interview Question

Write a Java program to check whether a number is a Perfect Number.

Understanding the Logic

Suppose

Number = 28

Find all divisors except

28

Add them together.

If the sum equals the original number,

then it is a Perfect Number.

Otherwise,

it is not.

Visual Representation

Input

28

Processing

Start

↓

Find Divisors

↓

1

↓

2

↓

4

↓

7

↓

14

↓

Sum

↓

1 + 2 + 4 + 7 + 14

↓

28

↓

Compare

↓

28 == 28

↓

Perfect Number

Brute Force Approach

The simplest solution is

  • Initialize the sum as 0.
  • Check every number from 1 to number - 1.
  • If it divides the number exactly, add it to the sum.
  • Compare the final sum with the original number.

Algorithm

Step 1

Read the number.

Step 2

Initialize

sum = 0;

Step 3

Loop from

1

to

number - 1

Step 4

Check

number % i == 0

Step 5

If true,

add

i

to

sum

Step 6

Compare

sum == number

Step 7

Print the result.

Dry Run

Input

6

Initial

sum = 0
i Divides 6? Sum
1 Yes 1
2 Yes 3
3 Yes 6
4 No 6
5 No 6

Comparison

6 == 6

Output

Perfect Number

Another Dry Run

Input

12

Initial

sum = 0
i Divides 12? Sum
1 Yes 1
2 Yes 3
3 Yes 6
4 Yes 10
5 No 10
6 Yes 16
7 No 16
8 No 16
9 No 16
10 No 16
11 No 16

Comparison

16 ≠ 12

Output

Not a Perfect Number

Approach 1 — Using Iteration (Loop)

This is the most common beginner-friendly approach and is frequently asked in interviews.

Complete Java Program

public class PerfectNumber {

    public static void main(String[] args) {

        int number = 28;

        int sum = 0;

        for (int i = 1; i < number; i++) {

            if (number % i == 0) {

                sum += i;

            }

        }

        if (sum == number) {

            System.out.println(number + " is a Perfect Number");

        } else {

            System.out.println(number + " is NOT a Perfect Number");

        }

    }

}

Output

28 is a Perfect Number

Step-by-Step Code Explanation

Step 1

Declare the number.

int number = 28;

Current value

28

Step 2

Initialize the sum.

int sum = 0;

Initially

sum = 0

Step 3

Start the loop.

for (int i = 1; i < number; i++)

The loop checks every possible proper divisor.

Step 4

Check divisibility.

if (number % i == 0)

If the remainder is zero,

then

i

is a divisor.

Step 5

Add the divisor.

sum += i;

Example for

28

sum = 1

↓

3

↓

7

↓

14

↓

28

Step 6

Compare the sum.

if (sum == number)

If true,

the number is perfect.

Otherwise,

it is not.

Example Execution

Input

Number = 496

Proper Divisors

1

2

4

8

16

31

62

124

248

Sum

496

Output

496 is a Perfect Number

Input

Number = 20

Proper Divisors

1

2

4

5

10

Sum

22

Output

20 is NOT a Perfect Number

Why Does This Work?

The algorithm checks every possible proper divisor of the given number.

Whenever a divisor is found, it is added to the running total.

If the final sum equals the original number, then the number satisfies the mathematical definition of a Perfect Number.

Advantages of This Approach

  • Very easy to understand.
  • Simple implementation using loops.
  • No extra data structures required.
  • Ideal for beginners.
  • Frequently asked in coding interviews.

Drawbacks

Although this approach is straightforward, it checks every number from 1 to N−1, making it inefficient for large inputs.

Interviewers often ask follow-up questions such as:

  • Can you optimize this solution?
  • Why don't we need to check every number?
  • Can you stop at the square root of the number?
  • How can you reduce the time complexity?
  • Can you create a reusable method?
  • What is the time complexity of both approaches?

In Part 2, we'll cover:

  • Optimized Approach Using Square Root
  • Reusable Method
  • Time and Space Complexity
  • Comparison of Approaches
  • Common Interview Mistakes
  • Interview Follow-up Questions
  • Related Coding Problems
  • Key Takeaways
  • Interview Tips

Approach 2 — Optimized Approach (Using Square Root)

The brute-force solution checks every number from

1

to

N - 1

This is inefficient for large numbers.

A better solution is to check divisors only up to the square root of the number.

This reduces the number of iterations significantly.

Why Does Square Root Optimization Work?

Divisors always occur in pairs.

Example

28

Divisor pairs

1 × 28

2 × 14

4 × 7

Once we reach

√28 ≈ 5.29

all remaining divisors have already been discovered as pairs.

Instead of checking

1 → 27

we only check

1 → 5

Visual Representation

Input

28

1  ←→ 28

2  ←→ 14

4  ←→ 7

Only check

1

2

3

4

5

Add

1

2 + 14

4 + 7

Result

28

Algorithm

Step 1

Read the number.

Step 2

Initialize

sum = 1;

Since

1

is always a proper divisor for numbers greater than 1.

Step 3

Loop

2

to

√number

Step 4

If

number % i == 0

then

i

is a divisor.

Step 5

Add

i

Also add its paired divisor

number / i

if they are different.

Step 6

Compare

sum == number

Dry Run

Input

28

Initial

sum = 1
i Divides? Added Values Sum
2 Yes 2 + 14 17
3 No - 17
4 Yes 4 + 7 28
5 No - 28

Comparison

28 == 28

Output

Perfect Number

Java Program

public class PerfectNumberOptimized {

    public static void main(String[] args) {

        int number = 28;

        if (number <= 1) {
            System.out.println("Not a Perfect Number");
            return;
        }

        int sum = 1;

        for (int i = 2; i * i <= number; i++) {

            if (number % i == 0) {

                sum += i;

                if (i != number / i) {

                    sum += number / i;

                }

            }

        }

        if (sum == number) {

            System.out.println(number + " is a Perfect Number");

        } else {

            System.out.println(number + " is NOT a Perfect Number");

        }

    }

}

Output

28 is a Perfect Number

Approach 3 — Using a Reusable Method

Reusable methods improve

  • Code readability
  • Maintainability
  • Unit testing
  • Code reuse

Java Program

public class PerfectNumberMethod {

    static boolean isPerfect(int number) {

        if (number <= 1) {

            return false;

        }

        int sum = 1;

        for (int i = 2; i * i <= number; i++) {

            if (number % i == 0) {

                sum += i;

                if (i != number / i) {

                    sum += number / i;

                }

            }

        }

        return sum == number;

    }

    public static void main(String[] args) {

        int number = 496;

        if (isPerfect(number)) {

            System.out.println(number + " is a Perfect Number");

        } else {

            System.out.println(number + " is NOT a Perfect Number");

        }

    }

}

Output

496 is a Perfect Number

Time Complexity

Brute Force Approach

Operation Complexity
Time O(n)
Space O(1)

Optimized Approach

Operation Complexity
Time O(√n)
Space O(1)

Comparison of Approaches

Approach Time Space Recommended
Brute Force O(n) O(1) Good for Beginners
Square Root Optimization O(√n) O(1) ✅ Best for Interviews
Reusable Method O(√n) O(1) Production Ready

Common Mistakes

Mistake 1

Including the number itself.

Wrong

for (int i = 1; i <= number; i++)

Correct

for (int i = 1; i < number; i++)

or use the optimized approach.

Mistake 2

Starting the sum with

0

when using the optimized approach.

Correct

sum = 1;

because

1

is always a proper divisor (for numbers greater than 1).

Mistake 3

Ignoring paired divisors.

Wrong

sum += i;

Correct

sum += i;
sum += number / i;

Mistake 4

Adding the square root twice.

Example

36

The divisor

6

pairs with itself.

Correct

if (i != number / i)

before adding the paired divisor.

Mistake 5

Ignoring edge cases.

Examples

0

1

Negative Numbers

None of these are Perfect Numbers.

Handle them before processing.

Interview Follow-up Questions

Q1. What is a Perfect Number?

Q2. Why do divisor pairs help optimize the solution?

Q3. Why is the optimized solution O(√n)?

Q4. Can you list the first four Perfect Numbers?

Q5. Can you check whether every even number is perfect?

Q6. Can a Perfect Number be odd?

Q7. Write a reusable method for checking Perfect Numbers.

Q8. Print all Perfect Numbers in a given range.

Q9. Compare the brute-force and optimized approaches.

Q10. What are some real-world applications of divisor-based algorithms?

Related Coding Problems

  • Prime Number
  • Armstrong Number
  • Strong Number
  • Harshad Number
  • GCD (HCF)
  • LCM
  • Factors of a Number
  • Sum of Divisors

Key Takeaways

  • A Perfect Number equals the sum of all its proper divisors.
  • The first few Perfect Numbers are:
6

28

496

8128
  • The brute-force solution checks every possible divisor and runs in O(n) time.
  • The optimized solution checks divisors only up to √n, reducing the time complexity to O(√n).
  • Divisors always appear in pairs, making square-root optimization possible.
  • Handle edge cases such as 0, 1, and negative numbers before processing.
  • Using a reusable method makes the code cleaner and easier to test.

Interview Tip

If an interviewer asks:

"Write a Java program to check whether a number is a Perfect Number."

Start with the straightforward loop-based solution to demonstrate your understanding of divisors. After completing it, explain that checking every number is inefficient and introduce the square-root optimization. Mention that divisors occur in pairs, allowing you to check only up to √n, which improves the time complexity from O(n) to O(√n). Finally, discuss edge cases (0, 1, and negative numbers) and show how to encapsulate the logic in a reusable method for production-quality code.