CODEWITHSUNDEEP
cfunctionprime-factoring
utdlegend
utdlegend

Reputation: 43

A function to find the sum of prime factors of an integer number

I am not able to meet all the conditions of prime factors of a number so that I find the correct sum. 

    int sumPrime(number){


         int counter;
         int sum=0;
         for(counter=2;counter<=(number/2);counter++){
           if(number%counter==0){
              sum=counter;
              counter++;
              sum+=counter;


              }
           return sum;  

          }
    }

Upvotes: 0

Views: 18944

Answers (4)

Paschalyn Ukwuani
Paschalyn Ukwuani

Reputation: 7

let n = 22;
let sum = 0;

function sumOfPrimeFactors(n) {
  // Write your code here
  if(n === 2){
    return true;
  }
  if(n < 2 || isNaN(n) || (n%2===0)){
    return false;
  }
  for(let i = 3; i < n; i++){
    if(n%i === 0){
      return false;
    }
  }
  return true;
}

for(let i = 0; i <= n; i++){
  if(sumOfPrimeFactors(i) === true){
    if(n % i === 0){
      sum += i;
    }
  }
}
console.log(`The sum of the prime factors in 91 is: ${sum}`)

Upvotes: 0

alindt
alindt

Reputation: 692

An efficient algorithm to find the prime factors

Code:

#include <stdio.h>
#include <math.h>

int sumPrime(int n) {
    int sum = 0;
    printf ("%6d: ", n);

    // find all powers of 2
    while (n%2 == 0) {
        printf("%d+", 2);
        n /= 2;
        sum += 2;
    }

    // checking for factors up to sqrt(n) is sufficient
    for (int i = 3; i <= sqrt(n); i += 2) {
        // find all powers of i
        while (n%i == 0) {
            printf("%d+", i);
            n /= i;
            sum += i;
        }
    }

    // n is prime at this point
    if (n > 2) {
        printf ("%d", n);
        sum += n;
    }

    printf (" = %d\n", sum);
    return sum;
}

void main() {
    sumPrime(226);
    sumPrime(39);
    sumPrime(775);
}

Output:

   226: 2+113 = 115
    39: 3+13 = 16
   775: 5+5+31 = 41

Upvotes: 0

Dimuth Tharaka Menikgama
Dimuth Tharaka Menikgama

Reputation: 122

#include <stdio.h>

int main(){


    printf("sum is%i",sumCount(12));
    return 0;
}
int sumCount(int number){
int count,sum=0;
for(count=2;count<number;count++){
    if(isPrime(count)==1){
        printf("%i \n",count);
        sum+=count;
    }
}
return sum;
}
int isPrime(int number){
    int i;
    for (i=2;i<=number/2 ;i++){
        if(number%i==0){
        return 0;
    }
}
return 1;
}

Upvotes: 1

MayurK
MayurK

Reputation: 1957

I think the code is self explanatory. It works for number >= 2.

int sumPrime(int number){
     int factor = 2;
     int sum=0;

    while(1 != number){ //Repeat the loop till number becomes 1.
       if(number%factor==0){ //Check if factor divides the number.
          number /= factor; //If yes, update the number.
          sum+=factor; //Add factor to sum.
          printf("Factor [%d]\n", factor);
          continue;
        }
        factor++; //If the current number is not a factor, check the next number.
    }
    printf("Sum [%d]\n", sum);
    return sum;
}

Output for 12:

Factor [2]                                                                                                                                      
Factor [2]                                                                                                                                      
Factor [3]                                                                                                                                      
Sum [7]

Upvotes: 2

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