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++

// 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;
}

Java

// 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
}

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#

// 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
}

PHP

<?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
?>

Javascript

<script>

// javascript 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)
    {
        var rem = (N + K) % K;

        if (rem == 0)
            return N;
        else
            return N + K - rem;
    }
// Driver Code


var N = 45, K = 6;

document.write("Smallest number greater than or equal to " + N
  +"<br>that is divisible by " + K + " is " + findNum(N, K));


// This code contributed by shikhasingrajput
</script>

Output:

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

Time Complexity: O(1)

Auxiliary Space: O(1)