Wednesday, February 1, 2012

Problem 21

Problem Statement :Problem21
Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.


Approach:
  • Function getvalue will return the value of all divisors of a number 
  • Call it twice and and just check whether they are equal or not.
Solution:
    public class problem21 {
     public static int isPrime(int n1) {
      int num = n1;
      for (int i = 2; i <= Math.sqrt(num); i++) {
       if (num % i == 0)
        return i;
      }
      return 0;
     }
    
    
     static int getvalue(int num) {
      if (isPrime(num) == 0)
       return 1;
      int value = 0;
      for (int i = 1; i <= num / 2; i++) {
       if (num % i == 0)
        value = value + i;
      }
      return value;
     }
    
     public static void main(String args[]) {
      int n = 10000, sum = 0;
      for (int i = 2; i < n; i++) {
       int val = getvalue(i);
       int revval = getvalue(val);
       if (i == revval && val != i)
        sum = sum + revval;
      }
      System.out.println(sum);
     }
    }
    

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