Here you can find the source of sumOfProperDivisors(long num)
public static long sumOfProperDivisors(long num)
//package com.java2s; /**/*from ww w . j av a 2 s . c om*/ * * maer - Solutions to problems of Project Euler * Copyright (C) 2011, Sandeep Gupta * http://www.sangupta.com/projects/maer * * The file is licensed under the the Apache License, Version 2.0 * (the "License"); you may not use this file except in compliance with * the License. You may obtain a copy of the License at * * http://www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software * distributed under the License is distributed on an "AS IS" BASIS, * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. * * See the License for the specific language governing permissions and * limitations under the License. * */ public class Main { public static long sumOfProperDivisors(long num) { if (num == 1) { return 0; } long sum = 1; // as 1 is a divisor and we are considering proper divisors long root = (long) Math.sqrt((double) num); if (root * root == num) { // case that n is a perfect square sum += root; root -= 1; } if (isOdd(num)) { for (int i = 2; i <= root; i += 2) { if (num % i == 0) sum += i + num / i; } } else { // number is even for (int i = 2; i <= root; i += 1) { if (num % i == 0) sum += i + num / i; } } return sum; } /** * Tests whether the given number is odd or not, uses bit arithmetic. * * @param number * @return */ public static boolean isOdd(long number) { return (number & 1) == 1; } }