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Smallest number required to be added to M to make it divisible by N

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  • Last Updated : 09 Jun, 2021
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Given two positive integers M and N, the task is to calculate the smallest number that needs to be added to M, to make it divisible by N.

Examples:

Input: M = 6, N = 7
Output: 1
Explanation: 1 is the smallest number that can be added to 6 to make it divisible by 7.

Input: M = 100, N = 28
Output: 12

Approach: The idea is to find the smallest number greater than or equal to M, that is divisible by N, and then subtract M from it. To get the smallest multiple of N ≥ M, divide M + N by N. If the remainder is 0, then the value is M. Otherwise, the value is M + N – remainder.

Below is the implementation of the above approach:

C++




// C++ program for 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;
}
 
// Function to find the smallest
// number required to be added to
// to M to make it divisible by N
int findSmallest(int M, int N)
{
    // Stores the smallest multiple
    // of N, greater than or equal to M
    int x = findNum(M, N);
 
    // Return the result
    return x - M;
}
 
// Driver Code
int main()
{
    // Given Input
    int M = 100, N = 28;
 
    // Function Call
    cout << findSmallest(M, N);
 
    return 0;
}

Java




// Java program for the above approach
class GFG{
 
// Function to find the smallest
// number greater than or equal
// to N, that is divisible by k
public static int findNum(int N, int K)
{
    int rem = (N + K) % K;
 
    if (rem == 0)
        return N;
    else
        return N + K - rem;
}
 
// Function to find the smallest
// number required to be added to
// to M to make it divisible by N
public static int findSmallest(int M, int N)
{
     
    // Stores the smallest multiple
    // of N, greater than or equal to M
    int x = findNum(M, N);
 
    // Return the result
    return x - M;
}
 
// Driver Code
public static void main(String args[])
{
     
    // Given Input
    int M = 100, N = 28;
 
    // Function Call
    System.out.println(findSmallest(M, N));
}}
 
// This code is contributed by SoumikMondal

Python3




# Python3 program for 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
 
# Function to find the smallest
# number required to be added to
# to M to make it divisible by N
def findSmallest(M, N):
     
    # Stores the smallest multiple
    # of N, greater than or equal to M
    x = findNum(M, N)
 
    # Return the result
    return x - M
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    M = 100
    N = 28
 
    # Function Call
    print(findSmallest(M, N))
     
# This code is contributed by ipg2016107

C#




// C# program for the above approach
using System;
         
class GFG{
     
// Function to find the smallest
// number greater than or equal
// to N, that is divisible by k
public static int findNum(int N, int K)
{
    int rem = (N + K) % K;
  
    if (rem == 0)
        return N;
    else
        return N + K - rem;
}
  
// Function to find the smallest
// number required to be added to
// to M to make it divisible by N
public static int findSmallest(int M, int N)
{
     
    // Stores the smallest multiple
    // of N, greater than or equal to M
    int x = findNum(M, N);
  
    // Return the result
    return x - M;
}
     
// Driver Code
public static void Main()
{
     
    // Given Input
    int M = 100, N = 28;
  
    // Function Call
    Console.WriteLine(findSmallest(M, N));
}
}
 
// This code is contributed by susmitakundugoaldanga

Javascript




<script>
 
// Javascript program for 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;
}
 
// Function to find the smallest
// number required to be added to
// to M to make it divisible by N
function findSmallest(M, N)
{
     
    // Stores the smallest multiple
    // of N, greater than or equal to M
    var x = findNum(M, N);
 
    // Return the result
    return x - M;
}
 
// Driver Code
 
// Given Input
var M = 100, N = 28;
 
// Function Call
document.write(findSmallest(M, N));
 
// This code is contributed by itsok
 
</script>

Output: 

12

 

Time Complexity: O(1)
Auxiliary Space: O(1)

 


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