quark

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06 2012 档案

摘要:Problem Description By replacing the 1st digit of *3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime. By replacing the 3rd and 4th digits of 56**3 with t... 阅读全文
posted @ 2012-06-26 17:13 QuarkZ 阅读(374) 评论(0) 推荐(0) 编辑

摘要:Problem Description In the card game poker, a hand consists of five cards and are ranked, from lowest to highest, in the following way: High Card: Highest value card. One Pair: Two cards of the same... 阅读全文
posted @ 2012-06-26 11:59 QuarkZ 阅读(230) 评论(0) 推荐(0) 编辑

摘要:Problem DescriptionThere 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,nCr =n!r!(n-r)!,where r<=n, n! = n*(n-1)…*3*2*1, and 0! = 1.It is not until n = 23, that a value exceeds 阅读全文
posted @ 2012-06-21 12:18 QuarkZ 阅读(150) 评论(0) 推荐(0) 编辑

摘要:Problem Description It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order. Find the smallest positive integer, x, such that 2x, 3x, ... 阅读全文
posted @ 2012-06-21 10:56 QuarkZ 阅读(171) 评论(0) 推荐(0) 编辑

摘要:Problem Description The prime 41, can be written as the sum of six consecutive primes: 41 = 2 + 3 + 5 + 7 + 11 + 13 This is the longest sum of consecutive primes that adds to a prime below one-hundred... 阅读全文
posted @ 2012-06-20 11:04 QuarkZ 阅读(298) 评论(0) 推荐(0) 编辑

摘要:Problem DescriptionThe 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-, 阅读全文
posted @ 2012-06-19 17:10 QuarkZ 阅读(318) 评论(0) 推荐(0) 编辑

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