Web Reference: Problem 5 is the smallest number that can be divided by each of the numbers from to without any remainder. What is the smallest positive number that is evenly divisible divisible with no remainder by all of the numbers from to ? We can take the biggest prime factors from 11-20 and multiply them. 20 = 2^2 * 5. 19 = 19. 18 = 2 * 3^2. 17 = 17. 16 = 2^4. 15 = 3 * 5. 14 = 2 * 7. 13 = 13. 11 = 11. Answer = 11 * 13 * 7 * 5 * 2^4 * 17 * 3^2 * 19. Or we can continuously take the lcm, for example a = lcm (20,19), b = lcm (a,18), etc. Aug 6, 2015 · There are many ways to solve this problem. Since you're doing an early exercise from Project Euler, I suppose you'd like to develop an approach that helps you understand the basic Python constructs (as opposed to the "batteries included" approach with gcd and so on).
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