Check if LCM of array elements is divisible by a prime number or not
Last Updated :
23 Jul, 2025
Given an array and a number k, the task is to find if LCM of the array is divisible by k or not.
Examples :
Input : int[] a = {10, 20, 15, 25}
k = 3
Output : true
Input : int[] a = {24, 21, 45, 57, 36};
k = 23;
Output : false
One simple solution is to first find LCM of array elements, then check if LCM is divisible by k or not.
Here, without calculating the LCM of the number we can find that LCM of the array of number is divisible by a prime number k or not. If any number of the array is divisible by prime number k, then the LCM of the number is also divisible by prime number k.
C++
// C++ program to find LCM of
// array of number is divisible
// by a prime number k or not
#include<iostream>
using namespace std;
// Function to check any number of
// array is divisible by k or not
bool func(int a[], int k, int n)
{
// If any array element is divisible
// by k, then LCM of whole array
// should also be divisible.
for (int i = 0; i < n; i++)
if (a[i] % k == 0)
return true;
return false;
}
// Driver Code
int main()
{
int a[] = {14, 27, 38, 76, 84};
int k = 19;
bool res = func(a, k, 5);
if(res)
cout<<"true";
else
cout<<"false";
return 0;
}
// This code is contributed
// by Mr. Somesh Awasthi
Java
// Java program to find LCM of
// array of number is divisible
// by a prime number k or not
import java.lang.*;
import java.util.*;
class GFG
{
// Function to check any number
// of array is divisible by k or not
static boolean func( int a[], int k)
{
// If any array element is divisible
// by k, then LCM of whole array
// should also be divisible.
for (int i = 0; i < a.length; i++)
if (a[i] % k == 0)
return true;
return false;
}
// Driver Code
public static void main(String args[])
{
int[] a = {14, 27, 38, 76, 84};
int k = 19;
boolean res = func(a, k);
System.out.println(res);
}
}
Python 3
# Python 3 program to find LCM
# of array of number is divisible
# by a prime number k or not
# Function to check any number of
# array is divisible by k or not
def func( a, k, n) :
# If any array element is
# divisible by k, then LCM
# of whole array should also
# be divisible.
for i in range(0, n) :
if ( a[i] % k == 0):
return True
# Driver Code
a = [14, 27, 38, 76, 84]
k = 19
res = func(a, k, 5)
if(res) :
print("true")
else :
print("false")
# This code is contributed
# by Nikita Tiwari.
C#
// C# program to find LCM of array
// of number is divisible by a prime
// number k or not
using System;
class GFG
{
// Function to check any number of
// array is divisible by k or not
static bool func(int []a, int k)
{
// If any array element is
// divisible by k, then LCM
// of whole array should also
// be divisible.
for (int i = 0; i < a.Length; i++)
if (a[i] % k == 0)
return true;
return false;
}
// Driver code
public static void Main()
{
int []a = {14, 27, 38, 76, 84};
int k = 19;
bool res = func(a, k);
Console.Write(res);
}
}
// This code is contributed by nitin mittal.
PHP
<?php
// PHP program to find LCM of
// array of number is divisible
// by a prime number k or not
// Function to check any number of
// array is divisible by k or not
function func( $a, $k, $n)
{
// If any array element is divisible
// by k, then LCM of whole array
// should also be divisible.
for ($i = 0; $i < $n; $i++)
if ($a[$i] % $k == 0)
return true;
return false;
}
// Driver Code
$a = array(14, 27, 38, 76, 84);
$k = 19;
$res = func($a, $k, 5);
if($res)
echo"true";
else
echo"false";
// This code is contributed by nitin mittal
?>
JavaScript
<script>
// javascript program to find LCM of
// array of number is divisible
// by a prime number k or not
// Function to check any number
// of array is divisible by k or not
function func(a , k)
{
// If any array element is divisible
// by k, then LCM of whole array
// should also be divisible.
for (let i = 0; i < a.length; i++)
if (a[i] % k == 0)
return true;
return false;
}
// Driver Code
let a = [ 14, 27, 38, 76, 84 ];
var k = 19;
let res = func(a, k);
document.write(res);
// This code is contributed by todaysgaurav
</script>
Output :
true
Time Complexity: O(n)
Auxiliary Space: O(1)
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