Check whether a number is Sublime number or not

Given a number N, the task is to check whether the number is a Sublime number or not. 

Sublime numbers are defined as positive numbers whose sum of positive divisors and count of positive divisors are both perfect numbers.

Examples:

Input: N = 12
Output: True
Explanation: 12 is a sublime number because its divisors are 1, 2, 3, 4, 6, and 12. So there are 6 factors of it and 6 is a perfect number and 1+2+3+4+6+12=28, which is also a perfect number. So 12 is a Sublime number.

Input: N = 24
Output: False
Explanation: 24 is a not sublime number because the divisors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. So there are 8 factors and which is not a perfect number and 1+2+3+4+6+8+12+24=60, Which is not a perfect number. So 24 is not a Sublime number.

Approach: We have to implement two main logic:

• Find and Count Divisors of a Number Logic
• Perfect Number Checking.

Below are the steps involved in the implementation of the code:

• Initialize a number n.
• Find the divisors of the given number (n).
• Count the total number of divisors of the given number (n) and store it in a variable (count).
• Sum up the divisors, and store the result in a variable (sum).
• At last, check whether count and sum are perfect numbers or not.
• If both are perfect numbers, the given number (n) is a sublime number.
• Else, not a sublime number.

Below is the implementation of the code:

C++

// C++ Implementation to Check whether
// a number is Sublime Number or Not.
#include <iostream>
using namespace std;

// Check whether the number is
// sublime or not
void checkSublime(int num)
{

    int s = 0, f = 0, s1 = 0, s2 = 0, i, j;

    for (i = 1; i <= num; i++) {
        if (num % i == 0) {

            // Finds the divisors
            // and sum up them
            s = s + i;

            // Counts the number
            // of divisors
            f++;
        }
    }

    // Check for perfect number
    for (j = 1; j < s; j++) {
        if (s % j == 0)
            s1 = s1 + j;
    }

    for (j = 1; j < f; j++) {
        if (f % j == 0)
            s2 = s2 + j;
    }

    if (s1 == s && s2 == f)
        cout << "True" << endl;
    else
        cout << "False" << endl;
}

// Driver code
int main()
{
    int n = 12;

    // Function call
    checkSublime(n);

    return 0;
}

Java

// java program to Check whether the number is sublime or
// not
public class SublimeNumber {

    // Check whether the number is sublime or not
    public static void checkSublime(int num)
    {
        int s = 0, f = 0, s1 = 0, s2 = 0, i, j;

        for (i = 1; i <= num; i++) {
            if (num % i == 0) {
                // Finds the divisors and sums them up
                s = s + i;
                // Counts the number of divisors
                f++;
            }
        }

        // Check for perfect number
        for (j = 1; j < s; j++) {
            if (s % j == 0)
                s1 = s1 + j;
        }

        for (j = 1; j < f; j++) {
            if (f % j == 0)
                s2 = s2 + j;
        }

        if (s1 == s && s2 == f)
            System.out.println("True");
        else
            System.out.println("False");
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 12;

        // Function call
        checkSublime(n);
    }
}

// This code is contributed by shivamgupta0987654321

Python3

# python3 Implementation to Check whether
# a number is Sublime Number or Not.

def checkSublime(num):
    s = 0
    f = 0
    s1 = 0
    s2 = 0

    for i in range(1, num+1):
        if num % i == 0:
            # Finds the divisors and sums them up
            s += i
            # Counts the number of divisors
            f += 1

    # Check for perfect number
    for j in range(1, s):
        if s % j == 0:
            s1 += j

    for j in range(1, f):
        if f % j == 0:
            s2 += j

    if s1 == s and s2 == f:
        print("True")
    else:
        print("False")


# Driver code
if __name__ == "__main__":
    n = 12

    # Function call
    checkSublime(n)


# This code is contributed by shivamgupta310570

C#

// C# Implementation to Check whether
// a number is Sublime Number or Not.
using System;

public class SublimeNumber
{
    // Check whether the number is sublime or not
    public static void CheckSublime(int num)
    {
        int s = 0, f = 0, s1 = 0, s2 = 0, i, j;

        for (i = 1; i <= num; i++)
        {
            if (num % i == 0)
            {
                // Finds the divisors and sums them up
                s = s + i;
                // Counts the number of divisors
                f++;
            }
        }

        // Check for perfect number
        for (j = 1; j < s; j++)
        {
            if (s % j == 0)
                s1 = s1 + j;
        }

        for (j = 1; j < f; j++)
        {
            if (f % j == 0)
                s2 = s2 + j;
        }

        if (s1 == s && s2 == f)
            Console.WriteLine("True");
        else
            Console.WriteLine("False");
    }

    // Driver code
    public static void Main(string[] args)
    {
        int n = 12;

        // Function call
        CheckSublime(n);
    }
}

Javascript

// Check whether the number is
// sublime or not
function checkSublime(num) {

    let s = 0, f = 0, s1 = 0, s2 = 0;

    for (let i = 1; i <= num; i++) {
        if (num % i === 0) {

            // Finds the divisors
            // and sum up them
            s += i;

            // Counts the number
            // of divisors
            f++;
        }
    }

    // Check for perfect number
    for (let j = 1; j < s; j++) {
        if (s % j === 0)
            s1 += j;
    }

    for (let j = 1; j < f; j++) {
        if (f % j === 0)
            s2 += j;
    }

    if (s1 === s && s2 === f)
        console.log("True");
    else
        console.log("False");
}

// Driver code
    const n = 12;

    // Function call
    checkSublime(n);

Output

True

Time Complexity: O(N)
Auxiliary Space: O(1) as using constant variables