Given a number n discover the smallest number same divisible by every number 1 come n.Examples:Input : n = 4Output : 12Explanation : 12 is the smallest numbers divisible by every numbers from 1 to 4Input : n = 10Output : 2520Input : n = 20Output : 232792560

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If you observe carefully the ans have to be the LCM of the numbers 1 to n.To discover LCM of numbers from 1 come n –Initialize ans = 1.Iterate over all the number from i = 1 to ns = n.At the i’th iteration ans = LCM(1, 2, …….., i). This have the right to be done conveniently as LCM(1, 2, …., i) = LCM(ans, i).Thus at i’th iteration us just need to do –ans = LCM(ans, i) = ans * ns / gcd(ans, i) Note : In C++ code, the answer conveniently exceeds the creature limit, even the lengthy long limit.Below is the implementation of the logic.
Output :232792560The above solution functions fine because that a solitary input. Yet if we have multiple inputs, that is a good idea to usage Sieve that Eratosthenes to store all prime factors. You re welcome refer below write-up for Sieve based approach.LCM of first n herbal NumbersThis article is contributed by Ayush Khanduri. If you prefer and would like to contribute, friend can also write an short article using or mail your write-up to contribute
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