Project Euler: Problem 21

By | 3. April 2015

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.

public class Problem21 {
	
	public static void main(String[] args) {
		int sum = 0;
		for (int i = 1; i < 10000; i++) {
			int d = d(i);
			if((d(d) == i && i != d)){
				sum += i;
				System.out.println(i);
			}
		}
		System.out.println(sum);
	}

	public static int d (int number){
		int sum = 0;
		for (int i = 1; i < number; i++) {
			if (number%i==0) {
				sum += i;
			}
		}
		return sum;
	}
	
}

Lösung: 31626

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