# Smallest number divisible by first n numbers

Given a number n find the smallest number evenly divisible by each number 1 to n.

Examples:

```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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 –

1. Initialize ans = 1.
2. 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.

## C++

```// C++ program to find smallest number evenly divisible by
// all numbers 1 to n
#include<bits/stdc++.h>
using namespace std;

// Function returns the lcm of first n numbers
long long lcm(long long n)
{
long long ans = 1;
for (long long i = 1; i <= n; i++)
ans = (ans * i)/(__gcd(ans, i));
return ans;
}

// Driver program to test the above function
int main()
{
long long n = 20;
cout << lcm(n);
return 0;
}
```

## Python

```# Python program to find the smallest number evenly
# divisible by all number 1 to n
import fractions

# Returns the lcm of first n numbers
def lcm(n):
ans = 1
for i in range(1, n + 1):
ans = (ans * i)/fractions.gcd(ans, i)
return ans

# main
n = 20
print lcm(n)
```

Output :

```232792560
```

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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