Project Euler
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Problem32
<br />We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.<br /><br /> The product 7254 is unusual, as the identity, 39*186 = 7254, con原创 2011-06-01 16:52:00 · 330 阅读 · 0 评论 -
Problem56
原题:A googol (power(10,100)) is a massive number: one followed by one-hundred zeros; power(100,100) is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of t原创 2011-06-22 14:53:00 · 265 阅读 · 0 评论 -
Problem58
原题:Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed. 37 36 35 34 33 32 31 38 17 16 15 14 13 30 39 18 5 4 3 12 29 40 19 6 1 2原创 2011-06-23 16:22:00 · 336 阅读 · 0 评论 -
Problem 46
原题: //合数是与质数(也叫素数)相对应的概念,即除了1和本身之外,还可以被其它正整数整除的数称为合数。 //奇合数:既是合数又是奇数的数称为奇合数。例如:9,15,21等等都是奇合数。 It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a原创 2011-06-14 17:15:00 · 316 阅读 · 0 评论 -
Problem45
原题:Triangle, pentagonal, and hexagonal numbers are generated by the following formulae: * Triangle T(n)=n(n+1)/2 1, 3, 6, 10, 15, ... Pentagonal P(n)=n(3n-1)/2 1, 5, 12, 22, 35, ... Hexagon原创 2011-06-13 15:33:00 · 328 阅读 · 0 评论 -
Problem44
原题:Pentagonal numbers are generated by the formula, Pn=n(3n-1)/2. The first ten pentagonal numbers are: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ... It can be seen that P4 + P7 = 22 + 70 = 92 = P8.原创 2011-06-13 15:32:00 · 342 阅读 · 0 评论 -
Problem41
原题:We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime. What is the largest n-digit原创 2011-06-08 17:48:00 · 333 阅读 · 0 评论 -
Problem40
原题:An irrational decimal(无理小数) fraction is created by concatenating the positive integers: 0.123456789101112131415161718192021... It can be seen that the 12(th) digit of the fractional part is 1. If原创 2011-06-08 14:14:00 · 272 阅读 · 0 评论 -
Problem42
原题:The nth term of the sequence of triangle numbers is given by, t(n) = 1/2*n(n+1); so the first ten triangle numbers are: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... By converting each letter in a wo原创 2011-06-09 16:38:00 · 311 阅读 · 0 评论 -
Problem39
原题:If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.{20,48,52}, {24,45,51}, {30,40,50} For which value of p分析:由原创 2011-06-08 13:32:00 · 287 阅读 · 0 评论 -
Problem38
原题:Take the number 192 and multiply it by each of 1, 2, and 3: 192*1 = 192 192*2 = 384 192*3 = 576 By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the原创 2011-06-08 08:52:00 · 314 阅读 · 0 评论 -
Problem37
原题:The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly w原创 2011-06-08 08:50:00 · 245 阅读 · 0 评论 -
Project Euler 36
原题:The decimal number, 585 = 1001001001 (binary), is palindromic in both bases. Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2. (Please note that the原创 2011-06-03 09:11:00 · 435 阅读 · 0 评论 -
Problem 28
Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows: 21 22 23 24 25 20 7 8 9 10 19 6 1 2 11 18 5 4 3 12 17 16 15 14 13 It can be verified that the sum of the numbers on th原创 2011-05-31 15:54:00 · 382 阅读 · 0 评论 -
Problem 29
<br />package cn.javass.mytest.java.euler; import java.util.Set; import java.util.TreeSet; /** * a(b) 表示 a的b次方 * * Consider all integer combinations of ab for 2<=a<=5 and 2<=b<=5: 2(2)=4, 2(3)=8, 2(4)=16, 2(5)=32 3(2)=9, 3(3)=27,原创 2011-05-31 15:46:00 · 257 阅读 · 0 评论 -
Project euler 34
原题:145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.Find the sum of all numbers which are equal to the sum of the factorial of their digits. Note: as 1! = 1 and 2! = 2 are not sums they原创 2011-06-03 09:07:00 · 300 阅读 · 0 评论 -
Project Euler 35
原题:The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime. There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31,原创 2011-06-03 09:09:00 · 393 阅读 · 0 评论 -
Problem57
原题:It is possible to show that the square root of two can be expressed as an infinite continued fraction. 2(开方) = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213... By expanding this for the first fou原创 2011-06-22 14:54:00 · 399 阅读 · 0 评论
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