LCM of N numbers modulo M

Given an array arr[] of integers, the task is to find the LCM of all the elements of the array modulo M where M = 109 + 7.

Examples:

Input: arr[] = {10000000, 12345, 159873}
Output: 780789722
LCM of (10000000, 12345, 159873) is 1315754790000000
1315754790000000 % 1000000007 = 780789722



Input: arr[] = {10, 13, 15}
Output: 390

Approach: If you have gone through the post that calculates LCM of array elements, the first approach that comes to mind is to take modulo at every step when LCM of ans and arr[i] is being calculated.

ans = 1
// For i = 1 to n – 1
ans = lcm(ans, arr[i]) % 1000000007 // Wrong approach

However, this approach is wrong and the mistake can be realized in the following example:

Take M = 41 and arr[] = {13, 18, 30}

Incorrect solution:
LCM(13, 18, 30) % 41
LCM(LCM(13, 18) % 41, 30) % 41
LCM(234 % 41, 30) % 41
LCM(29, 30) % 41
870 % 41
9

Correct solution:
LCM(13, 18, 30) % 41
LCM(LCM(13, 18), 30) % 41
LCM(234, 30) % 41
1170 % 41
22

Note: Whenever the LCM of 2 numbers becomes > M, the approach doesn’t work.

The correct approach is to prime factorize the elements of the array and keep track of the highest power of every prime for each element. LCM will be the product of these primes raised to their highest power in the array.

Illustration:


Let elements be [36, 480, 500, 343]
Prime factorization results:
36 = 22 * 32
480 = 25 * 3 * 5
500 = 22 * 53
343 = 73

Highest power of 2 amongt all array elements = Max(2, 5, 2, 0) = 5
Highest power of 3 amongt all array elements = Max(2, 1, 0, 0) = 2
Highest power of 5 amongt all array elements = Max(0, 1, 3, 0) = 3
Highest power of 7 amongt all array elements = Max(0, 0, 0, 3) = 3

Therefore, LCM = 25 * 32 * 53 * 73 = 12348000

Let p be a prime factor of an element of the array and x be its highest power in the whole array. Then,

 LCM = \prod_{p}p^x

Using the above formula, we can easily calculate LCM of the whole array and our problem of MOD will also be solved. Simplifying the expression, we get:

 LCM = p_1^{x_1} * p_2^{x_2} * ... * p_n^{x_n}

Since modulo operation is distributive over multiplication, we can safely write the following expression.

 LCM = (((p_1^{x_1} * p_2^{x_2}) \% MOD) * p_3^{x_3}) \% MOD \hspace{0.2cm} and \hspace{0.2cm} so \hspace{0.2cm} on..

Now, the problem arises as to how to compute prime factors and their powers efficiently. For that, we can use the sieve of Eratosthenes. Refer to this post: Using Sieve to compute prime factors and their powers.


Below is the implementation of the above approach:

C++

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// C++ program to compute LCM of array elements modulo M
#include <bits/stdc++.h>
#define F first
#define S second
#define MAX 10000003
using namespace std;
  
typedef long long ll;
const int mod = 1000000007;
  
int prime[MAX];
unordered_map<int, int> max_map;
  
// Function to return a^n
int power(int a, int n)
{
    if (n == 0)
        return 1;
    int p = power(a, n / 2) % mod;
    p = (p * p) % mod;
    if (n & 1)
        p = (p * a) % mod;
    return p;
}
  
// Function to find the smallest prime factors
// of numbers upto MAX
void sieve()
{
    prime[0] = prime[1] = 1;
    for (int i = 2; i < MAX; i++) {
        if (prime[i] == 0) {
            for (int j = i * 2; j < MAX; j += i) {
                if (prime[j] == 0) {
                    prime[j] = i;
                }
            }
            prime[i] = i;
        }
    }
}
  
// Function to return the LCM modulo M
ll lcmModuloM(const int* ar, int n)
{
  
    for (int i = 0; i < n; i++) {
        int num = ar[i];
        unordered_map<int, int> temp;
  
        // Temp stores mapping of prime factor to
        // its power for the current element
        while (num > 1) {
  
            // Factor is the smallest prime factor of num
            int factor = prime[num];
  
            // Increase count of factor in temp
            temp[factor]++;
  
            // Reduce num by its prime factor
            num /= factor;
        }
  
        for (auto it : temp) {
  
            // Store the highest power of every prime
            // found till now in a new map max_map
            max_map[it.first] = max(max_map[it.first], it.second);
        }
    }
  
    ll ans = 1;
  
    for (auto it : max_map) {
  
        // LCM is product of primes to their highest powers modulo M
        ans = (ans * power(it.F, it.S)) % mod;
    }
  
    return ans;
}
  
// Driver code
int main()
{
    sieve();
    int arr[] = { 36, 500, 480, 343 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << lcmModuloM(arr, n);
    return 0;
}

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Python3

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# Python3 program to compute LCM of 
# array elements modulo M
MAX = 10000003
  
mod = 1000000007
  
prime = [0 for i in range(MAX)]
  
max_map = dict()
  
# function to return a^n
def power(a, n):
      
    if n == 0:
        return 1
    p = power(a, n // 2) % mod
    p = (p * p) % mod
      
    if n & 1:
        p = (p * a) % mod
    return p
  
# function to find the smallest prime
# factors of numbers upto MAX
def sieve():
    prime[0], prime[1] = 1, 1
    for i in range(2, MAX):
        if prime[i] == 0:
            for j in range(i * 2, MAX, i):
                if prime[j] == 0:
                    prime[j] = i
            prime[i] = i
  
# function to return the LCM modulo M
def lcmModuloM(arr, n):
      
    for i in range(n):
        num = arr[i]
          
        temp = dict()
          
        # temp stores mapping of prime factors 
        # to its power for the current element
        while num > 1:
               
            # factor is the smallest prime
            # factor of num
            factor = prime[num]
              
            # Increase count of factor in temp
            if factor in temp.keys():
                temp[factor] += 1
            else:
                temp[factor] = 1
                  
            # Reduce num by its prime factor
            num = num // factor
              
        for i in temp:
            # store the higest power of every prime
            # found till now in a new map max_map
            if i in max_map.keys():
                max_map[i] = max(max_map[i], temp[i])
            else:
                max_map[i] = temp[i]
              
    ans = 1
      
    for i in max_map:
          
        # LCM is product of primes to their
        # higest powers modulo M
        ans = (ans * power(i, max_map[i])) % mod
    return ans
  
# Driver code
sieve()
arr = [36, 500, 480, 343]
n = len(arr)
print(lcmModuloM(arr, n))
  
# This code is contributed
# by Mohit kumar 29

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Output:

12348000

The above code works for the following constraints:

 1 <= N <= 10^6 \newline 1 <= A[i] <= 10^7

References: https://stackoverflow.com/questions/16633449/calculate-lcm-of-n-numbers-modulo-1000000007



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