Smallest number greater than or equal to N divisible by K

Given a number N and a number K, the task is to find the smallest number greater than or equal to N which is divisible by K.

Examples:

Input: N = 45, K = 6
Output: 48
48 is the smallest number greater than or equal to 45
which is divisible by 6.

Input: N = 11, K = 3
Output: 12


Approach: The idea is to divide the N+K by K. If the remainder is 0 then print N else print N + K – remainder.

Below is the implementation of the above approach :

C++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the smallest number
// greater than or equal to N
// that is divisible by k
int findNum(int N, int K)
{
    int rem = (N + K) % K;
  
    if (rem == 0)
        return N;
    else
        return N + K - rem;
}
  
// Driver code
int main()
{
    int N = 45, K = 6;
  
    cout << "Smallest number greater than or equal to " << N
         << "\nthat is divisible by " << K << " is " << findNum(N, K);
  
    return 0;
}

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Java

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// Java implementation of the above approach
  
public class GFG{
  
    // Function to find the smallest number
    // greater than or equal to N
    // that is divisible by k
    static int findNum(int N, int K)
    {
        int rem = (N + K) % K;
      
        if (rem == 0)
            return N;
        else
            return N + K - rem;
    }
  
  
     // Driver Code
     public static void main(String []args){
           
        int N = 45, K = 6;
  
    System.out.println("Smallest number greater than or equal to " + N
          +"\nthat is divisible by " + K + " is " + findNum(N, K));
  
     }
     // This code is contributed by ANKITRAI1
}

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Python

# Python 3 implementation of the
# above approach

# Function to find the smallest number
# greater than or equal to N
# that is divisible by k
def findNum(N, K):
rem = (N + K) % K;

if (rem == 0):
return N
else:
return (N + K – rem)

# Driver Code
N = 45
K = 6
print(‘Smallest number greater than’,
‘or equal to’ , N,
‘that is divisible by’, K,
‘is’ , findNum(45, 6))

# This code is contributed by Arnab Kundu

C#

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// C# implementation of the above approach
  
public class GFG{
  
    // Function to find the smallest number
    // greater than or equal to N
    // that is divisible by k
    static int findNum(int N, int K)
    {
        int rem = (N + K) % K;
      
        if (rem == 0)
            return N;
        else
            return N + K - rem;
    }
  
  
    // Driver Code
    static void Main(){
          
        int N = 45, K = 6;
  
    System.Console.WriteLine("Smallest number greater than or equal to " + N
        +"\nthat is divisible by " + K + " is " + findNum(N, K));
  
    }
    // This code is contributed by mits
}

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PHP

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<?php
// PHP implementation of the above approach
  
// Function to find the smallest number
// greater than or equal to N that is
// divisible by k
function findNum($N, $K)
{
    $rem = ($N + $K) % $K;
  
    if ($rem == 0)
        return $N;
    else
        return $N + $K - $rem;
}
  
// Driver code
$N = 45; $K = 6;
  
echo "Smallest number greater than "
                   "or equal to ", $N;
echo "\nthat is divisible by " , $K
            " is " , findNum($N, $K);
  
// This code is contributed by anuj_67
?>

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Output:

Smallest number greater than or equal to 45
that is divisible by 6 is 48


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