Smallest number greater than or equal to N divisible by K

Last Updated : 13 Apr, 2021

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes