This problem is a programming version of Problem 1 from projecteuler.net. 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. \$\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. Solution Obvious solution. Ask Question Asked today. Project Euler Problem 1 Statement. Since you have a generalized solution, you should provide your generalized specification of the problem. I solve Project Euler problems to practice and extend my math and program­ming skills, all while having fun at the same time. Project Euler (projecteuler.net) is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Please be sure to answer the question. If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. Make an attempt, then come back with specific questions about any difficulties you're having. Here I make my solutions publicly available for other enthusiasts to learn from and to critique. We won't just give you a solution - what would you learn if we did? – Fabinout Jan 29 '14 at 16:24 The sum of these multiples is 23. The first challenge is the following. If we list all the natural numbers below that are multiples of or , we get and . May 3, 2011 Programming C++, Code, Project Euler Rian. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and … If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. … 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. I'm learning C++ and currently doing Project Euler challenges. The sum of these multiples is 23. – Tangentially Perpendicular 15 mins ago Find the sum of all the multiples of 3 or 5 below 1000. Especially when somebody already pointed out the problem in her code. Project Euler #1: Multiples of 3 and 5. 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. Problem. ... but it was the Project Euler forums that showed me the formula to do it. Project Euler solutions Introduction. The sum of these multiples is 23. Project Euler is a teaching resource. Problem 48 of Project Euler has the nice and simple description. 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. I suppose that Euler_Project are for getting better at programming, I don't think you should handle answers as easily as you did. 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