mu chance or much chance ?

日々の戯れ言

CLASS極に向けて

rb

海の日ですがゲーセンで遊んでいます.CLASS極に向けて,全譜面AAAを目指します.まずはスキルレート85以下AAAを目指そうかと.MISSも無くさないと.

帰国

深夜便で帰国します.長いようであっという間の出張でした.

マクドナルド

ホテルでは英語が通じたのですが,マクドナルドでは英語が通じなくて,出張中一番焦りました.

バラ色の街

休憩時間の合間にトゥールーズを観光しました. キャピトル広場 サン・セルナン・バジリカ聖堂 次に行く機会があれば,一日中観光したい.

パン料理

やはりパン料理に飽きました.飽きたと言いながらハンバーガーを食べていますが・・・.そろそろ日本食を食べたい・・・.

フレンチフライ

フレンチフライが食べたくて,ハンバーガープレートを頂きました.まだメニューの英語を見ても,どのようなものが出てくるかは想像できません.

ロッシーニ風

食べる機会がありましたので,頂きました.フォアグラ美味しかったです.

パエリア

食べる機会がありましたので,頂きました.美味しかったですが,食べづらかったです.

時差ボケ

やっぱり時差ボケに悩まされますね.いい加減慣れないと・・・.

フランス

出張でフランスに行きます.今回で2回目です.タクシーとホテルで英語が通じたので何とかなりそうです.

Project Euler 50

問題 Problem 50:Consecutive prime sum 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.The longest sum of c…

Project Euler 49

問題 Problem 49:Prime permutations 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 permutati…

Project Euler 48

問題 Problem 48:Self powers The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000. 解答例 sum = 0 for i in range(1, 1001): sum += i ** i tempStr = str(sum) print(…

Project Euler 47

問題 Problem 47:Distinct primes factors The first two consecutive numbers to have two distinct prime factors are:14 = 2 × 7 15 = 3 × 5The first three consecutive numbers to have three distinct prime factors are:644 = 2^2 × 7 × 23 645 = 3 …

Project Euler 46

問題 Problem 46:Goldbach's other conjecture 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 + 2×1^2 15 = 7 + 2×2^2 21 = 3 + 2×3^2 25 = 7 + 2×3^2 27 = 19 +…

Project Euler 45

問題 Problem 45:Triangular, pentagonal, and hexagonal Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:Triangle:Tn=n(n+1)/2 1, 3, 6, 10, 15, ... Pentagonal:Pn=n(3n−1)/2 1, 5, 12, 22, 35, ... Hexagonal…

Project Euler 44

問題 Problem 44:Pentagon numbers 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. However, thei…

Project Euler 43

問題 Problem 43:Sub-string divisibility The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.Let d1 b…

Project Euler 42

問題 Problem 42:Coded triangle numbers 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 numb…

Project Euler 41

問題 Problem 41:Pandigital prime 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 p…

Project Euler 40

問題 Problem 40:Champernowne's constant An irrational decimal fraction is created by concatenating the positive integers:0.123456789101112131415161718192021...It can be seen that the 12th digit of the fractional part is 1.If dn represents…

Project Euler 39

問題 Problem 39:Integer right triangles 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 ≤ 1000, is t…

Project Euler 38

問題 Problem 38:Pandigital multiples 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 co…

Project Euler 37

問題 Problem 37:Truncatable primes 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 wo…

Project Euler 36

問題 Problem 36:Double-base palindromes 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 pal…

Project Euler 35

問題 Problem 35:Circular primes 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,…

Project Euler 34

問題 Problem 34:Digit factorials 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 in…

Project Euler 33

問題 Problem 33:Digit cancelling fractions The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the …

Project Euler 32

問題 Problem 32:Pandigital products 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.The product 7254 is unusual, as…

Project Euler 31

問題 Problem 31:Coin sums In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p). It is possible to make £2 in the following way:1×…