Admirable Numbers
Last Updated :
13 Jul, 2021
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++
#include <bits/stdc++.h>
using namespace std;
int divSum( int n)
{
int result = 0;
for ( int i = 2; i <= sqrt (n); i++) {
if (n % i == 0) {
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
return (result + n + 1);
}
bool check( int num)
{
int sigmaN = divSum(num);
for ( int i = 2; i <= sqrt (num); i++) {
if (num % i == 0) {
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 ;
}
}
}
if (sigmaN - 2 * 1 == 2 * num)
return true ;
return false ;
}
bool isAdmirableNum( int N)
{
return check(N);
}
int main()
{
int n = 12;
if (isAdmirableNum(n))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}
|
Java
class GFG{
static int divSum( int n)
{
int result = 0 ;
for ( int i = 2 ; i <= Math.sqrt(n); i++)
{
if (n % i == 0 )
{
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
return (result + n + 1 );
}
static boolean check( int num)
{
int sigmaN = divSum(num);
for ( int i = 2 ; i <= Math.sqrt(num); i++)
{
if (num % i == 0 )
{
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 ;
}
}
}
if (sigmaN - 2 * 1 == 2 * num)
return true ;
return false ;
}
static boolean isAdmirableNum( int N)
{
return check(N);
}
public static void main(String[] args)
{
int n = 12 ;
if (isAdmirableNum(n))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Python3
import math
def divSum(n):
result = 0
for i in range ( 2 , int (math.sqrt(n)) + 1 ):
if (n % i = = 0 ):
if (i = = (n / / i)):
result + = i
else :
result + = (i + n / / i)
return (result + n + 1 )
def check(num):
sigmaN = divSum(num)
for i in range ( 2 , int (math.sqrt(num)) + 1 ):
if (num % i = = 0 ):
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
if (sigmaN - 2 * 1 = = 2 * num):
return True
return False
def isAdmirableNum(N):
return check(N)
n = 12
if (isAdmirableNum(n)):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG{
static int divSum( int n)
{
int result = 0;
for ( int i = 2; i <= Math.Sqrt(n); i++)
{
if (n % i == 0)
{
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
return (result + n + 1);
}
static bool check( int num)
{
int sigmaN = divSum(num);
for ( int i = 2; i <= Math.Sqrt(num); i++)
{
if (num % i == 0)
{
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 ;
}
}
}
if (sigmaN - 2 * 1 == 2 * num)
return true ;
return false ;
}
static bool isAdmirableNum( int N)
{
return check(N);
}
public static void Main()
{
int n = 12;
if (isAdmirableNum(n))
Console.Write( "Yes" );
else
Console.Write( "No" );
}
}
|
Javascript
<script>
function divSum( n)
{
let result = 0;
for ( let i = 2; i <= Math.sqrt(n); i++)
{
if (n % i == 0)
{
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
return (result + n + 1);
}
function check( num) {
let sigmaN = divSum(num);
for (let i = 2; i <= Math.sqrt(num); i++) {
if (num % i == 0) {
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 ;
}
}
}
if (sigmaN - 2 * 1 == 2 * num)
return true ;
return false ;
}
function isAdmirableNum( N) {
return check(N);
}
let n = 12;
if (isAdmirableNum(n))
document.write( "Yes" );
else
document.write( "No" );
</script>
|
Time Complexity: O(N1/2)
References: OEIS
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