Given a number n uncover the smallest number evenly divisible by each number 1 to n.Examples:Input : n = 4Output : 12Explacountry : 12 is the smallest numbers divisible by all numbers from 1 to 4Input : n = 10Output : 2520Input : n = 20Output : 232792560

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If you observe very closely the ans need to be the LCM of the numbers 1 to n.To uncover LCM of numbers from 1 to n –Initialize ans = 1.Iterate over all the numbers from i = 1 to i = n.At the i’th iteration ans = LCM(1, 2, …….., i). This deserve to be done quickly as LCM(1, 2, …., i) = LCM(ans, i).Hence at i’th iteration we simply have to execute –ans = LCM(ans, i) = ans * i / gcd(ans, i) Note : In C++ code, the answer conveniently exceeds the integer limit, also the long long limit.Below is the implementation of the logic.
Output :232792560The over solution works fine for a solitary input. But if we have multiple inputs, it is an excellent concept to use Sieve of Eratosthenes to store all prime determinants. Please refer listed below short article for Sieve based technique.LCM of First n Natural NumbersThis short article is added by Ayush Khanduri. If you favor and also would choose to contribute, you deserve to additionally create an short article using or mail your article to contribute See your short article appearing on the major web page and also help other Geeks.Attention reader! Don’t sheight learning currently. Get organize of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To finish your preparation from learning a language to DS Algo and also many kind of even more, please refer Complete Interwatch Preparation Course.

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