Sum of all proper divisors of a natural number
Last Updated :
23 Jun, 2022
Given a natural number, calculate sum of all its proper divisors. A proper divisor of a natural number is the divisor that is strictly less than the number.
For example, number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.
Examples :
Input : num = 10
Output: 8
// proper divisors 1 + 2 + 5 = 8
Input : num = 36
Output: 55
// proper divisors 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 = 55
This problem has very simple solution, we all know that for any number ‘num’ all its divisors are always less than and equal to ‘num/2’ and all prime factors are always less than and equal to sqrt(num). So we iterate through ‘i’ till i<=sqrt(num) and for any ‘i’ if it divides ‘num’ , then we get two divisors ‘i’ and ‘num/i’ , continuously add these divisors but for some numbers divisors ‘i’ and ‘num/i’ will same in this case just add only one divisor , e.g; num=36 so for i=6 we will get (num/i)=6 , that’s why we will at 6 in the summation only once. Finally we add one as one is divisor of all natural numbers.
C++
#include<bits/stdc++.h>
using namespace std;
int divSum( int num)
{
int result = 0;
if (num == 1)
return result;
for ( int i=2; i<= sqrt (num); i++)
{
if (num%i==0)
{
if (i==(num/i))
result += i;
else
result += (i + num/i);
}
}
return (result + 1);
}
int main()
{
int num = 36;
cout << divSum(num);
return 0;
}
|
Java
import java.math.*;
class GFG {
static int divSum( int num)
{
int result = 0 ;
for ( int i = 2 ; i <= Math.sqrt(num); i++)
{
if (num % i == 0 )
{
if (i == (num / i))
result += i;
else
result += (i + num / i);
}
}
return (result + 1 );
}
public static void main(String[] args)
{
int num = 36 ;
System.out.println(divSum(num));
}
}
|
Python3
import math
def divSum(num) :
result = 0
i = 2
while i< = (math.sqrt(num)) :
if (num % i = = 0 ) :
if (i = = (num / i)) :
result = result + i;
else :
result = result + (i + num / i);
i = i + 1
return (result + 1 );
num = 36
print (divSum(num))
|
C#
using System;
class GFG {
static int divSum( int num)
{
int result = 0;
for ( int i = 2; i <= Math.Sqrt(num); i++)
{
if (num % i == 0)
{
if (i == (num / i))
result += i;
else
result += (i + num / i);
}
}
return (result + 1);
}
public static void Main()
{
int num = 36;
Console.Write(divSum(num));
}
}
|
PHP
<?php
function divSum( $num )
{
$result = 0;
for ( $i = 2; $i <= sqrt( $num );
$i ++)
{
if ( $num % $i == 0)
{
if ( $i == ( $num / $i ))
$result += $i ;
else
$result += ( $i + $num / $i );
}
}
return ( $result + 1);
}
$num = 36;
echo (divSum( $num ));
?>
|
Javascript
<script>
function divSum(num)
{
let result = 0;
for (let i=2; i<=Math.sqrt(num); i++)
{
if (num%i==0)
{
if (i==(num/i))
result += i;
else
result += (i + num/i);
}
}
return (result + 1);
}
let num = 36;
document.write(divSum(num));
</script>
|
Output :
55
Time Complexity: O(?n)
Auxiliary Space: O(1)
Please refer below post for an optimized solution and formula.
Efficient solution for sum of all the factors of a number
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