Java Program to Check if a Given Number is Perfect Number

Last Updated : 30 Jul, 2023

A number is said to be a perfect number if the sum of its proper divisors ( i.e. all positive divisors excluding the number itself )is equal to that number itself. Aliquot sum is the sum of divisors of a number, excluding the number itself. Hence, a number is a perfect number only if it is equal to its aliquot sum. All known perfect numbers are even. In this article, we will learn how to Check Perfect Numbers in Java.

Example 1:

n = 9
Proper Divisors of 9 are 1 and 3.
Sum = 1+3 = 4 ≠ 9
⇒ 9 is not a perfect number.

Example 2:

n = 6
Proper Divisors of 6 are 1, 2 and 3.
Sum = 1+2+3 = 6 = 6
⇒ 6 is a perfect number

So, we basically have to find the sum of the proper divisors of a number.

1. Using Loop to Check Perfect Number in Java

A Simple Solution is to go through every number from 1 to n-1 and check if it is a divisor and if it is, then add it in the sum variable and at the end check if the sum is equal to the number itself, then it is a perfect number otherwise not.

Below is the implementation of the above method:

Java

// Java program to check if a given // number is perfect or not  class GFG {      // Returns true if n is perfect     static boolean isPerfect(int n)     {         // 1 is not a perfect number         if (n == 1)             return false;          // sum will store the sum of proper divisors         // As 1 is a proper divisor for all numbers         // initialised sum with 1         int sum = 1;          // Looping through the numbers to check if they are         // divisors or not         for (int i = 2; i < n; i++) {              if (n % i == 0) {                 sum += i;             }         }          // If sum of divisors is equal to         // n, then n is a perfect number         if (sum == n)             return true;          return false;     }      // Driver program     public static void main(String[] args)     {         int n = 6;          // Call isPerfect function to         // check if the number is perfect or not.         if (isPerfect(n))             System.out.println(n + " is a perfect number");         else             System.out.println(                 n + " is not a perfect number");     } }

Output

6 is a perfect number

The complexity of the above method

Time Complexity: O(n)

2. Using Square root to Check Perfect Number in Java

An Efficient Solution is to go through numbers till the square root of n.

If i is a divisor then n/i is also a divisor.

Java

// Java program to check if a given // number is perfect or not  class GFG {      // Returns true if n is perfect     static boolean isPerfect(int n)     {         // 1 is not a perfect number         if (n == 1)             return false;          // sum will store the sum of proper divisors         // As 1 is a proper divisor for all numbers         // initialised sum with 1         int sum = 1;          // Looping through the numbers to check if they are         // divisors or not         for (int i = 2; i * i <= n; i++) {              if (n % i == 0) {                  // n is a perfect square                 // let's take 25                 // we need to add 5 only once                 // sum += i + n / i will add it twice                  if (i * i == n)                     sum += i;                 else                     sum += i + (n / i);             }         }          // If sum of divisors is equal to         // n, then n is a perfect number          if (sum == n)              return true;          return false;     }      // Driver program     public static void main(String[] args)     {         int n = 6;          // Call isPerfect function to         // check if the number is perfect or not.         if (isPerfect(n))              System.out.println(n + " is a perfect number");          else             System.out.println(                 n + " is not a perfect number");     } }

Output

6 is a perfect number

The complexity of the above method

Time Complexity: O(√n)

3. Recursive Approach to Check Perfect Number in Java

Below is the implement the above method:

Java

// Java Program to implement Perfect // Number using Recursion import java.util.*;  // Driver Class public class GFG {     static long sum = 0;      static long isPerfect(long num, int i)     {         // Base Condition         if (i <= num / 2) {             if (num % i == 0) {                 sum = sum + i;             }              // after each iteration, increments the value of             // variable i by 1             i++;              // recursive call             isPerfect(num, i);         }         // returns the sum of factors         return sum;     }      // main function     public static void main(String args[])     {         long number = 28, s;         int i = 1;          s = isPerfect(number, i);          // compares sum with the number         if (s == number)             // prints if the s and number are equal             System.out.println(number                                + " is a perfect number");         else             // prints if s and number are not equal             System.out.println(                 number + " is not a perfect number");     } }

Output

28 is a perfect number

The complexity of the above method

Time Complexity: O(n)

Java Program to Check if a Given Integer is Positive or Negative

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Java Program to Check if a Given Number is Perfect Number

1. Using Loop to Check Perfect Number in Java

Below is the implementation of the above method:

The complexity of the above method

2. Using Square root to Check Perfect Number in Java

The complexity of the above method

3. Recursive Approach to Check Perfect Number in Java

The complexity of the above method

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