随笔分类 - 欧拉计划
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：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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