题意: 完全数是指真因数之和等于自身的那些数.例如,28的真因数之和为1 + 2 + 4 + 7 + 14 = 28,因此28是一个完全数. 一个数n被称为亏数,如果它的真因数之和小于n:反之则被称为盈数. 由于12是最小的盈数,它的真因数之和为1 + 2 + 3 + 4 + 6 = 16,所以最小的能够表示成两个盈数之和的数是24.通过数学分析可以得出,所有大于28123的数都可以被写成两个盈数的和:尽管我们知道最大的不能被写成两个盈数的和的数要小于这个值,但这是通过分析所能得到的最好上界.…

Special subset sums: testing 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: S(B) ≠ S(C); that is, sums of subsets ca…

本题来自 Project Euler 第12题:https://projecteuler.net/problem=12 # Project Euler: Problem 12: Highly divisible triangular number # The sequence of triangle numbers is generated by adding the natural numbers. # So the 7th triangle number would be 1 + 2 + 3…

上一次接触 project euler 还是2011年的事情,做了前三道题,后来被第四题卡住了,前面几题的代码也没有保留下来. 今天试着暴力破解了一下,代码如下: (我大概是第 172,719 个解出这道题的人) program 4 A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.…

开始做 Project Euler 的练习题.网站上总共有565题,真是个大题库啊! # Project Euler, Problem 1: Multiples of 3 and 5 # If we list all the natural numbers below 10 # that are multiples of 3 or 5, we get 3, 5, 6 and 9. # The sum of these multiples is 23. # Find the sum of all…

Project Euler: 欧拉计划是一系列挑战数学或者计算机编程问题,解决这些问题需要的不仅仅是数学功底. 启动这一项目的目的在于,为乐于探索的人提供一个钻研其他领域并且学习新知识的平台,将这一平台打造一个有趣和休闲 的环境. 项目主页:https://projecteuler.net 第一题 Multiples of 3 and 5 If we list all the natural numbers below 10 that are multiples of 3 or 5, we ge…

本题来自 Project Euler 第17题:https://projecteuler.net/problem=17 ''' Project Euler 17: Number letter counts If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total. If all th…

本题来自 Project Euler 第11题:https://projecteuler.net/problem=11 # Project Euler: Problem 10: Largest product in a grid # In the 20×20 grid below, four numbers along a diagonal line have been marked in red. # The product of these numbers is 26 × 63 × 78 ×…

题意:三个正整数a + b + c = 1000,a*a + b*b = c*c.求a*b*c. 解法:可以暴力枚举,但是也有数学方法. 首先,a,b,c中肯定有至少一个为偶数,否则和不可能为以上两个等式均不会成立.然后,不可能a,b为奇c为偶,否则a*a%4=1, b*b%4=1, 有(a*a+b*b) %4 = 2,而c*c%4 = 0.也就是说,a和b中至少有一个偶数. 这是勾股数的一个性质,a,b中至少有一个偶数. 然后,解决过程见下(来自project euler的讨论): tag:m…

In Problem 42 we dealt with triangular problems, in Problem 44 of Project Euler we deal with pentagonal number, I can only wonder if we have to deal with septagonal numbers in Problem 46. Anyway the problem reads Pentagonal numbers are generated by t…