02 2014 档案
摘要：Consider the fraction,n/d, wherenanddare positive integers. Ifndand HCF(n,d)=1, it is called a reduced proper fraction.If we list the set of reduced proper fractions ford8 in ascending order of size, we get:1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3,3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6,
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摘要：Thenthterm of the sequence of triangle numbers is given by,tn= ½n(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 word to a number corresponding to its alphabetical position and adding these values we form a word value. For exam
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摘要：We shall say that ann-digit number is pandigital if it makes use of all the digits 1 tonexactly once. For example, 2143 is a 4-digit pandigital and is also prime.What is the largestn-digit pandigital prime that exists?题目大意：如果一个数字将1到n的每个数字都使用且只使用了一次，我们将其称其为一个n位的pandigital数。例如，2143是一个4位的pandigital数，并且
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摘要：Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of positive numbers less than or equal tonwhich are relatively prime ton. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6. The number 1 is conside
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摘要：The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:1! + 4! + 5! = 1 + 24 + 120 = 145Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it turns out that there are only three such loops that ex
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摘要：It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.9 = 7 + 21215 = 7 + 22221 = 3 + 23225 = 7 + 23227 = 19 + 22233 = 31 + 212It turns out that the conjecture was false.What is the smallest odd composite that cannot be written
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摘要：The cube, 41063625 (3453), can be permuted to produce two other cubes: 56623104 (3843) and 66430125 (4053). In fact, 41063625 is the smallest cube which has exactly three permutations of its digits which are also cube.Find the smallest cube for which exactly five permutations of its digits are cube.
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摘要：Consider the fraction,n/d, wherenanddare positive integers. Ifndand HCF(n,d)=1, it is called a reduced proper fraction.If we list the set of reduced proper fractions ford8 in ascending order of size, we get:1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6
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摘要：There are exactly ten ways of selecting three from five, 12345:123, 124, 125, 134, 135, 145, 234, 235, 245, and 345In combinatorics, we use the notation,5C3= 10.In general,It is not untiln= 23, that a value exceeds one-million:23C10= 1144066.How many, not necessarily distinct, values of nCr, for 1n1
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摘要：The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes,
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摘要：The first two consecutive numbers to have two distinct prime factors are:14 = 27 15 = 35The first three consecutive numbers to have three distinct prime factors are:644 = 2²723 645 = 3543 646 = 21719.Find the first four consecutive integers to have four distinct prime factors. What is the first
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摘要：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 we can work from right to left: 3797, 379, 37, and 3.Find the sum of the only eleven primes that are both
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摘要：The decimal number, 585 = 10010010012(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 palindromic number, in either base, may not include leading zeros.)题目大意：十进制数字585 = 10010010012(二进制),可以看出在十
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摘要：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, 37, 71, 73, 79, and 97.How many circular primes are there below one million?题目大意：我们称197为一个循环质数，因为它的所有轮转形式: 19
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摘要：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 are not included.题目大意：145 是一个奇怪的数字, 因为 1! + 4! + 5! = 1 + 24 + 120 = 145.找出所有等于各位数字阶乘之和的数字之和。注意: 因为 1! =
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摘要：The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that49/98 = 4/8, which is correct, is obtained by cancelling the 9s.We shall consider fractions like, 30/50 = 3/5, to be trivial examples.There are exactly four non-trivia
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摘要：Consider all integer combinations ofabfor 2a5 and 2b5:22=4, 23=8, 24=16, 25=3232=9, 33=27, 34=81, 35=24342=16, 43=64, 44=256, 45=102452=25, 53=125, 54=625, 55=3125If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:4, 8, 9, 16, 25,
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摘要：Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:2122 23 242520 7 89 1019 6 1 2 1118 5 4 3 121716 15 1413It can be verified that the sum of the numbers on the diagonals is 101.What is the sum of the numbers on the diagonals in a 1001 ..
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摘要：Usingnames.txt(right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a na
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摘要：It is possible to show that the square root of two can be expressed as an infinite continued fraction.2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...By expanding this for the first four iterations, we get:1 + 1/2 = 3/2 = 1.51 + 1/(2 + 1/2) = 7/5 = 1.41 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...1
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摘要：Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n). If d(a) = b and d(b) = a, where a b, then a and b are an amicable pair and each of a and b are called amicable numbers.For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44,.
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摘要：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 the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?NOTE:Do not count spaces or hyphens. For examp
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摘要：215= 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.What is the sum of the digits of the number 21000?题目大意：题目大意：215= 32768 并且其各位之和为 is 3 + 2 + 7 + 6 + 8 = 26.21000的各位数之和是多少？// （Problem 16）Power digit sum// Completed on Sun, 17 Nov 2013, 15:23// Language: C//// 版权所有（C）acutus (mail: acutu..
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摘要：Starting in the top left corner of a 22 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner.How many such routes are there through a 2020 grid?题目大意：从一个22网格的左上角开始，有6条（不允许往回走）通往右下角的路。对于2020的网格，这样的路有多少条？// （Problem 15）Lattice paths// Completed
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摘要：The following iterative sequence is defined for the set of positive integers:n n/2 (n is even) n 3n + 1 (n is odd)Using the rule above and starting with 13, we generate the following sequence:13 40 20 10 5 16 8 4 2 1It can be seen that this sequence (starting at 13 and finishing at 1) con...
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摘要：Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.371072875339021027987979982208375902465101357402504637693767749000971264812489697007805041701826053874324986199524741059474233309513058123726617309629919422133635741615725224305633018110724061549082502306758820753
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摘要：The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.Find the sum of all the primes below two million.#include#include#include#define N 2000000bool prim(int n){ int i; for(i=2; i*i<=n; i++) { if(n%i==0) return false; } return true;}int main(){ int i; long long ...
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摘要：A Pythagorean triplet is a set of three natural numbers,abc, for which,a2+b2=c2For example, 32+ 42= 9 + 16 = 25 = 52.There exists exactly one Pythagorean triplet for whicha+b+c= 1000.Find the productabc.#include#include#include#include#include#includevoid show(){ int a,b,c; for(a=1; a<333; a++...
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摘要：By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.What is the 10 001st prime number?#include #include #include #include int prim(int n){ int i; for(i=2; i*i<=n; i++) { if(n%i==0) return 0; } return 1;} void solve(int n){ int...
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摘要：Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025385 = 2640.Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.前十个自然数的平方和是：12+ 22+ ... + 102= 385前十个自然数的和的平方是：(1 + 2
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摘要：2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.What is the smallest positive number that isevenly divisibleby all of the numbers from 1 to 20?#include #include #include #include #define N 20 int gcd(int a, int b){ if(b==0) return a; ...
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摘要：A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 9199.Find the largest palindrome made from the product of two 3-digit numbers.#include#include#include#include#include#includebool palindromic(int n) //判断一个整数是否为回文数{ char ...
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摘要：The prime factors of 13195 are 5, 7, 13 and 29.What is the largest prime factor of the number 600851475143 ?#include#include#include#include#include#include#define N 600851475143bool prim(int n){ int i; for(i=2; i*i<=n; i++) { if(n%i==0) return false; } return true;}...
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摘要：Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-value
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摘要：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 the multiples of 3 or 5 below 1000.题目大意：10以下的自然数中，属于3和5的倍数的有3,5,6和9，它们之和是23.找出1000以下的自然数中，属于3和5的倍数的数字之和。#include #include #include void solve(){ in...
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