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)