/* 
 * Solution to Project Euler problem 88
 * Copyright (c) Project Nayuki. All rights reserved.
 * 
 * https://www.nayuki.io/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;


public final class p088 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p088().run());
	}
	
	
	private static final int LIMIT = 12000;
	
	
	/* 
	 * minSumProduct[k] is the smallest positive integers that can be written as both a sum and a product of the same collection of k positive integers.
	 * For example, minSumProduct[3] = 6 because 6 = 1 + 2 + 3 = 1 * 2 * 3, and this is the minimum possible number for 3 terms.
	 * 
	 * For all k >= 2:
	 * - minSumProduct[k] > k because 1 + ... + 1 (with k terms) = k, which is the minimum sum of k positive integers,
	 *   but the product is 1 which is unequal to k, so k is not a valid solution.
	 * - minSumProduct[k] <= 2k because 1 + ... + 1 + 2 + k (with k terms in total) = (k - 2) + 2 + k = 2k. The product is 2k, which equals the sum.
	 *   Since this is one achievable solution, the minimum solution must be no larger than this.
	 * - Aside: minSumProduct[k] is not a prime number. Suppose minSumProduct[k] = p, where p is prime. Then p can only be factorized as p, p * 1, p * 1 * 1, etc.
	 *   So whenever the factorization has more than one term, the sum exceeds p, which makes it unequal to the product.
	 * 
	 * Therefore we need to consider all numbers from 2 to LIMIT*2 and factorize them in all possible ways to find all the relevant solutions.
	 */
	private int[] minSumProduct;
	
	
	public String run() {
		minSumProduct = new int[LIMIT + 1];
		Arrays.fill(minSumProduct, Integer.MAX_VALUE);
		for (int i = 2; i <= LIMIT * 2; i++)
			factorize(i, i, i, 0, 0);
		
		// Eliminate duplicates and compute sum
		Set<Integer> items = new HashSet<>();
		for (int i = 2; i < minSumProduct.length; i++)
			items.add(minSumProduct[i]);
		int sum = 0;
		for (int n : items)
			sum += n;
		return Integer.toString(sum);
	}
	
	
	/* 
	 * Calculates all factorizations of the integer n >= 2 and updates smaller solutions into minSumProduct.
	 * For example, 12 can be factorized as follows - and duplicates are eliminated by finding only non-increasing sequences of factors:
	 * - 12 = 12. (1 term)
	 * - 12 = 6 * 2 * 1 * 1 * 1 * 1 = 6 + 2 + 1 + 1 + 1 + 1. (6 terms)
	 * - 12 = 4 * 3 * 1 * 1 * 1 * 1 * 1 = 4 + 3 + 1 + 1 + 1 + 1 + 1. (7 terms)
	 * - 12 = 3 * 2 * 2 * 1 * 1 * 1 * 1 * 1 = 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1. (8 terms)
	 */
	private void factorize(int n, int remain, int maxFactor, int sum, int terms) {
		if (remain == 1) {
			if (sum > n)  // Without using factors of 1, the sum never exceeds the product
				throw new AssertionError();
			
			terms += n - sum;
			if (terms <= LIMIT && n < minSumProduct[terms])
				minSumProduct[terms] = n;
			
		} else {
			// Note: maxFactor <= remain
			for (int i = 2; i <= maxFactor; i++) {
				if (remain % i == 0) {
					int factor = i;
					factorize(n, remain / factor, Math.min(factor, maxFactor), sum + factor, terms + 1);
				}
			}
		}
	}
	
}