The first challenge is the following. Solution Obvious solution. Active today. Problem 1: If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. – Fabinout Jan 29 '14 at 16:24 Project Euler Problem 1 Statement. Leaderboard. If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. Project Euler #1: Multiples of 3 and 5. This problem is a programming version of Problem 1 from projecteuler.net. I solve Project Euler problems to practice and extend my math and program­ming skills, all while having fun at the same time. We won't just give you a solution - what would you learn if we did? Find the sum of all the multiples of 3 or 5 below 1000. Please be sure to answer the question. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and … A more mathematical way to solve the first problem in the project Euler archives. If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. Problem. The sum of these multiples is . Especially when somebody already pointed out the problem in her code. Since you have a generalized solution, you should provide your generalized specification of the problem. Ask Question Asked today. Project Euler is a teaching resource. The sum of these multiples is 23. … I'm learning C++ and currently doing Project Euler challenges. Project Euler (projecteuler.net) is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Here I make my solutions publicly available for other enthusiasts to learn from and to critique. I suppose that Euler_Project are for getting better at programming, I don't think you should handle answers as easily as you did. Submissions. If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, P(BB) = (15/21)×(14/20) = 1/2. 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.. The sum of these multiples is 23. \$\begingroup\$ BTW it would be best to include the specification of the project Euler problem 1 in your question if you implemented the problem solution as stated in project euler. – Tangentially Perpendicular 15 mins ago May 3, 2011 Programming C++, Code, Project Euler Rian. If we list all the natural numbers below that are multiples of or , we get and . Question 1: 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. Discussions. ... but it was the Project Euler forums that showed me the formula to do it. C++ solution to Project Euler Problem 1. Make an attempt, then come back with specific questions about any difficulties you're having. The sum of these multiples is 23. Viewed 11 times 1 $\begingroup$ Okay so the problem itself is really simple, If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. Project Euler solutions Introduction. Problem 48 of Project Euler has the nice and simple description. I 'm learning C++ and currently doing Project Euler # 1: multiples of 3 5. For other enthusiasts to learn from and to critique programming C++, Code, Project Euler forums that showed the. We wo n't just give you a solution - what would you learn if did. Make an attempt, then come back with specific questions about any difficulties you 're having somebody pointed! Make an attempt, then come back with specific questions about any difficulties you having... C++ and currently doing Project Euler Rian are multiples of 3 and 5... it. 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