# 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 = 10 ^{9} + 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

**is being calculated.**

*arr[i]*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

9Correct 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 = 2^{2}* 3^{2}

480 = 2^{5}* 3 * 5

500 = 2^{2}* 5^{3}

343 = 7^{3}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) = 3Therefore, LCM = 2

^{5}* 3^{2}* 5^{3}* 7^{3}= 12348000

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

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:

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

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++

`// 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; ` `} ` |

*chevron_right*

*filter_none*

## Python3

# 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

**Output:**

12348000

The above code works for the following constraints:

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

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