text: Sum of all proper divisors of a natural number | C # program to find sum of all divisorsof a natural number ; Function to calculate sum of all proper divisors num -- > given natural number ; Final result of summation of divisors ; find all divisors which divides ' num ' ; if ' i ' is divisor of ' num ' ; if both divisors are same then add it only once else add both ; Add 1 to the result as 1 is also a divisor ; Driver Code
code: 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 ) ) ; } }