## Problem:

Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:

1. S(B) [≠] S(C); that is, sums of subsets cannot be equal.
2. If B contains more elements than C then S(B) [>] S(C).

If S(A) is minimised for a given n, we shall call it an optimum special sum set. The first five optimum special sum sets are given below.

n = 1: {1}
n = 2: {1, 2}
n = 3: {2, 3, 4}
n = 4: {3, 5, 6, 7}
n = 5: {6, 9, 11, 12, 13}

It seems that for a given optimum set, A = {a1, a2, ... , an}, the next optimum set is of the form B = {b, a1+b, a2+b, ... ,an+b}, where b is the "middle" element on the previous row.

By applying this "rule" we would expect the optimum set for n = 6 to be A = {11, 17, 20, 22, 23, 24}, with S(A) = 117. However, this is not the optimum set, as we have merely applied an algorithm to provide a near optimum set. The optimum set for n = 6 is A = {11, 18, 19, 20, 22, 25}, with S(A) = 115 and corresponding set string: 111819202225.

Given that A is an optimum special sum set for n = 7, find its set string.

NOTE: This problem is related to problems 105 and 106.

6857

## Code:The solution may include methods that will be found here: Library.java .

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 3 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */public final class p003 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p003().run());	}			/* 	 * Algorithm: Divide out all the smallest prime factors except the last one.	 * For example, 1596 = 2 * 2 * 3 * 7 * 19. The algorithm ensures that the smallest factors will be found first.	 * After dividing out the smallest factors, the last factor to be found will be equal to the quotient, so it must be the largest prime factor.	 */	public String run() {		long n = 600851475143L;		while (true) {			long p = smallestFactor(n);			if (p < n)				n /= p;			else				return Long.toString(n);		}	}			private static long smallestFactor(long n) {		for (long i = 2, end = Library.sqrt(n); i <= end; i++) {			if (n % i == 0)				return i;		}		return n;  // Prime	}	}