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














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# 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 



















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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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