Java Program to Find sum of even factors of a number
Last Updated :
03 Nov, 2022
Given a number n, the task is to find the even factor sum of a number. Examples:
Input : 30
Output : 48
Even dividers sum 2 + 6 + 10 + 30 = 48
Input : 18
Output : 26
Even dividers sum 2 + 6 + 18 = 26
Let p1, p2, … pk be prime factors of n. Let a1, a2, .. ak be highest powers of p1, p2, .. pk respectively that divide n, i.e., we can write n as n = (p1a1)*(p2a2)* … (pkak).
Sum of divisors = (1 + p1 + p12 ... p1a1) *
(1 + p2 + p22 ... p2a2) *
...........................
(1 + pk + pk2 ... pkak)
If number is odd, then there are no even factors, so we simply return 0. If number is even, we use above formula. We only need to ignore 20. All other terms multiply to produce even factor sum. For example, consider n = 18. It can be written as 2132 and sum of all factors is (20 + 21)*(30 + 31 + 32). if we remove 20 then we get the Sum of even factors (2)*(1+3+32) = 26. To remove odd number in even factor, we ignore then 20 which is 1. After this step, we only get even factors. Note that 2 is the only even prime.
java
import java.util.*;
import java.lang.*;
public class GfG {
public static int sumofFactors( int n)
{
if (n % 2 != 0 )
return 0 ;
int res = 1 ;
for ( int i = 2 ; i <= Math.sqrt(n); i++) {
int count = 0 , curr_sum = 1 ;
int curr_term = 1 ;
while (n % i == 0 ) {
count++;
n = n / i;
if (i == 2 && count == 1 )
curr_sum = 0 ;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
if (n >= 2 )
res *= ( 1 + n);
return res;
}
public static void main(String argc[])
{
int n = 18 ;
System.out.println(sumofFactors(n));
}
}
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Please refer complete article on Find sum of even factors of a number for more details!
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