Admirable Numbers

Given an integer N, the task is to check if N is an Admirable Number.

An admirable number is a number, if there exists a proper divisor D’ of N such that sigma(N)-2D’ = 2N, where sigma(N) is the sum of all divisors of N 

Examples:

Input: N = 12 
Output: Yes 
Explanation:
12’s proper divisors are 1, 2, 3, 4, 6, and 12 
sigma(N) = 1 + 2 + 3 + 4 + 6 + 12 = 28 
sigma(N) – 2D’ = 2N 
28 – 2*2 = 2*12 
24 == 24 

Input: N = 28 
Output: No 

Approach: The idea is to find the sum of all factors of a number that is sigma(N). And then we will find every proper divisor of a number D’ and check if there exists a proper divisor D’ of N such that 

Below is the implementation of the above approach:

C++

// C++ implementation to check if
// N is an admirable number

#include <bits/stdc++.h>
using namespace std;

// Function to calculate the sum of
// all divisors of a given number
int divSum(int n)
{
    // Sum of divisors
    int result = 0;

    // Find all divisors
    // which divides 'num'
    for (int i = 2; i <= sqrt(n); i++) {

        // if 'i' is divisor of 'n'
        if (n % i == 0) {

            // if both divisors are same
            // then add it once else add
            if (i == (n / i))
                result += i;
            else
                result += (i + n / i);
        }
    }

    // Add 1 and n to result as above loop
    // considers proper divisors greater
    return (result + n + 1);
}

// Function to check if there
// exists a proper divisor
// D' of N such that sigma(n)-2D' = 2N
bool check(int num)
{
    int sigmaN = divSum(num);

    // Find all divisors which divides 'num'
    for (int i = 2; i <= sqrt(num); i++) {

        // if 'i' is divisor of 'num'
        if (num % i == 0) {

            // if both divisors are same then add
            // it only once else add both
            if (i == (num / i)) {
                if (sigmaN - 2 * i == 2 * num)
                    return true;
            }
            else {
                if (sigmaN - 2 * i == 2 * num)
                    return true;
                if (sigmaN - 2 * (num / i) == 2 * num)
                    return true;
            }
        }
    }

    // Check 1 since 1 is also a divisor
    if (sigmaN - 2 * 1 == 2 * num)
        return true;

    return false;
}

// Function to check if N
// is an admirable number
bool isAdmirableNum(int N)
{
    return check(N);
}

// Driver code
int main()
{
    int n = 12;
    if (isAdmirableNum(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java implementation to check if N
// is a admirable number
class GFG{

// Function to calculate the sum of
// all divisors of a given number
static int divSum(int n)
{
    // Sum of divisors
    int result = 0;

    // Find all divisors
    // which divides 'num'
    for (int i = 2; i <= Math.sqrt(n); i++)
    {

        // if 'i' is divisor of 'n'
        if (n % i == 0)
        {

            // if both divisors are same
            // then add it once else add
            if (i == (n / i))
                result += i;
            else
                result += (i + n / i);
        }
    }

    // Add 1 and n to result as above loop
    // considers proper divisors greater
    return (result + n + 1);
}

// Function to check if there
// exists a proper divisor
// D' of N such that sigma(n)-2D' = 2N
static boolean check(int num)
{
    int sigmaN = divSum(num);

    // Find all divisors which divides 'num'
    for (int i = 2; i <= Math.sqrt(num); i++)
    {

        // if 'i' is divisor of 'num'
        if (num % i == 0)
        {

            // if both divisors are same then add
            // it only once else add both
            if (i == (num / i))
            {
                if (sigmaN - 2 * i == 2 * num)
                    return true;
            }
            else
            {
                if (sigmaN - 2 * i == 2 * num)
                    return true;
                if (sigmaN - 2 * (num / i) == 2 * num)
                    return true;
            }
        }
    }

    // Check 1 since 1 is also a divisor
    if (sigmaN - 2 * 1 == 2 * num)
        return true;

    return false;
}

// Function to check if N
// is an admirable number
static boolean isAdmirableNum(int N)
{
    return check(N);
}

// Driver code
public static void main(String[] args)
{
    int n = 12;
    if (isAdmirableNum(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}

// This code is contributed by shubham

Python3

# Python3 implementation to check if
# N is an admirable number
import math

# Function to calculate the sum of
# all divisors of a given number
def divSum(n):

    # Sum of divisors
    result = 0

    # Find all divisors
    # which divides 'num'
    for i in range(2, int(math.sqrt(n)) + 1):

        # If 'i' is divisor of 'n'
        if (n % i == 0):

            # If both divisors are same
            # then add it once else add
            if (i == (n // i)):
                result += i
            else:
                result += (i + n // i)

    # Add 1 and n to result as above loop
    # considers proper divisors greater
    return (result + n + 1)

# Function to check if there
# exists a proper divisor
# D' of N such that sigma(n)-2D' = 2N
def check(num):

    sigmaN = divSum(num)

    # Find all divisors which divides 'num'
    for i in range(2, int(math.sqrt(num)) + 1):

        # If 'i' is divisor of 'num'
        if (num % i == 0):

            # If both divisors are same then add
            # it only once else add both
            if (i == (num // i)):
                if (sigmaN - 2 * i == 2 * num):
                    return True

            else:
                if (sigmaN - 2 * i == 2 * num):
                    return True
                if (sigmaN - 2 * (num // i) == 2 * num):
                    return True

    # Check 1 since 1 is also a divisor
    if (sigmaN - 2 * 1 == 2 * num):
        return True

    return False

# Function to check if N
# is an admirable number
def isAdmirableNum(N):

    return check(N)

# Driver code
n = 12

if (isAdmirableNum(n)):
    print("Yes")
else:
    print("No")

# This code is contributed by divyeshrabadiya07

C#

// C# implementation to check if N
// is a admirable number
using System;
class GFG{

// Function to calculate the sum of
// all divisors of a given number
static int divSum(int n)
{

    // Sum of divisors
    int result = 0;

    // Find all divisors
    // which divides 'num'
    for(int i = 2; i <= Math.Sqrt(n); i++)
    {

       // If 'i' is divisor of 'n'
       if (n % i == 0)
       {

           // If both divisors are same
           // then add it once else add
           if (i == (n / i))
               result += i;
           else
               result += (i + n / i);
       }
    }

    // Add 1 and n to result as above loop
    // considers proper divisors greater
    return (result + n + 1);
}

// Function to check if there
// exists a proper divisor
// D' of N such that sigma(n)-2D' = 2N
static bool check(int num)
{
    int sigmaN = divSum(num);

    // Find all divisors which divides 'num'
    for(int i = 2; i <= Math.Sqrt(num); i++)
    {

       // If 'i' is divisor of 'num'
       if (num % i == 0)
       {

           // If both divisors are same then add
           // it only once else add both
           if (i == (num / i))
           {
               if (sigmaN - 2 * i == 2 * num)
                   return true;
           }
           else
           {
               if (sigmaN - 2 * i == 2 * num)
                   return true;
               if (sigmaN - 2 * (num / i) == 2 * num)
                   return true;
           }
       }
    }

    // Check 1 since 1 is also a divisor
    if (sigmaN - 2 * 1 == 2 * num)
        return true;

    return false;
}

// Function to check if N
// is an admirable number
static bool isAdmirableNum(int N)
{
    return check(N);
}

// Driver code
public static void Main()
{
    int n = 12;

    if (isAdmirableNum(n))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by Code_Mech

Javascript

<script>
// Javascript implementation to check if N
// is a admirable number

    // Function to calculate the sum of
    // all divisors of a given number
    function divSum( n)
    {
        // Sum of divisors
        let result = 0;

        // Find all divisors
        // which divides 'num'
        for ( let i = 2; i <= Math.sqrt(n); i++)
        {

            // if 'i' is divisor of 'n'
            if (n % i == 0)
            {

                // if both divisors are same
                // then add it once else add
                if (i == (n / i))
                    result += i;
                else
                    result += (i + n / i);
            }
        }

        // Add 1 and n to result as above loop
        // considers proper divisors greater
        return (result + n + 1);
    }

    // Function to check if there
    // exists a proper divisor
    // D' of N such that sigma(n)-2D' = 2N
    function check( num) {
        let sigmaN = divSum(num);

        // Find all divisors which divides 'num'
        for (let i = 2; i <= Math.sqrt(num); i++) {

            // if 'i' is divisor of 'num'
            if (num % i == 0) {

                // if both divisors are same then add
                // it only once else add both
                if (i == (num / i)) {
                    if (sigmaN - 2 * i == 2 * num)
                        return true;
                } else {
                    if (sigmaN - 2 * i == 2 * num)
                        return true;
                    if (sigmaN - 2 * (num / i) == 2 * num)
                        return true;
                }
            }
        }

        // Check 1 since 1 is also a divisor
        if (sigmaN - 2 * 1 == 2 * num)
            return true;

        return false;
    }

    // Function to check if N
    // is an admirable number
    function isAdmirableNum( N) {
        return check(N);
    }

    // Driver code
    let n = 12;
    if (isAdmirableNum(n))
        document.write("Yes");
    else
        document.write("No");

// This code is contributed by todaysgaurav
</script>

Output:

Yes

Time Complexity: O(N^1/2)

References: OEIS