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 all numbers native 1 to 4Input : n = 10Output : 2520Input : n = 20Output : 232792560


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If you observe very closely the ans should be the LCM the the number 1 come n.To discover LCM of numbers from 1 to n –Initialize ans = 1.Iterate over every the numbers from i = 1 to ns = n.At the i’th iteration ans = LCM(1, 2, …….., i). This can be done quickly as LCM(1, 2, …., i) = LCM(ans, i).Thus in ~ i’th iteration we just have to do –ans = LCM(ans, i) = ans * i / gcd(ans, i) Note : In C++ code, the answer quickly exceeds the integer limit, even the long long limit.Below is the implementation the the logic.
Output :232792560The over solution functions fine because that a single input. However if we have actually multiple inputs, it is a good idea to use Sieve the Eratosthenes to store all element factors. You re welcome refer below post for Sieve based approach.LCM of first n organic NumbersThis short article is added by Ayush Khanduri. If you prefer barisalcity.org and would like to contribute, you can likewise write an write-up using contribute.barisalcity.org or letter your post to contribute
barisalcity.org. Check out your article appearing on the barisalcity.org main page and assist other Geeks.Attention reader! Don’t stop discovering now. Acquire hold of all the important mathematical concepts for vain programming through the Essential Maths because that CP Course in ~ a student-friendly price. To complete your preparation from finding out a language come DS Algo and many more, you re welcome refer Complete Interview preparation Course.


Count the number of pairs (i, j) such that either arr is divisible by arr or arr is divisible by arr
Kth element in permutation of first N organic numbers having actually all even numbers placed before odd number in raising order
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