(Problem 1)Multiples of 3 and 5】的更多相关文章

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…
Consider the fraction, n/d, where n and d are positive integers. If nd and HCF(n,d)=1, it is called a reduced proper fraction. If we list the set of reduced proper fractions for d  8 in ascending order of size, we get: 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1…
The nth term 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…
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 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…
Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of positive numbers less than or equal to n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relative…
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 = 145 Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it tu…
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 + 212 It turns out that the conjecture was false. What…
Consider the fraction, n/d, where n and d are positive integers. If nd and HCF(n,d)=1, it is called a reduced proper fraction. If we list the set of reduced proper fractions for d  8 in ascending order of size, we get: 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1…
There are exactly ten ways of selecting three from five, 12345: 123, 124, 125, 134, 135, 145, 234, 235, 245, and 345 In combinatorics, we use the notation, 5C3 = 10. In general, It is not until n = 23, that a value exceeds one-million: 23C10 = 114406…
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 seq…