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 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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