Given a number n find the smallest number evenly divisible by each number 1 to n.
Input : n = 4 Output : 12 Explanation : 12 is the smallest numbers divisible by all numbers from 1 to 4 Input : n = 10 Output : 2520 Input : n = 20 Output : 232792560
If you observe carefully the ans must be the LCM of the numbers 1 to n.
To find 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 can be done easily as LCM(1, 2, …., i) = LCM(ans, i).
Thus at i’th iteration we just have to do –
ans = LCM(ans, i) = ans * i / gcd(ans, i) [Using the below property, a*b = gcd(a,b) * lcm(a,b)]
Note : In C++ code, the answer quickly exceeds the integer limit, even the long long limit.
Below is the implementation of the logic.
The above solution works fine for a single input. But if we have multiple inputs, it is a good idea to use Sieve of Eratosthenes to store all prime factors. Please refer below article for Sieve based approach.
LCM of First n Natural Numbers
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