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:
Approach to solve this problem would be to start from N and check each number one by one until we find a number that is divisible by K. We can use a while loop to keep incrementing N until we find a number that is divisible by K.
Here are the steps of approach:
- We start with N and check each number one by one until we find a number that is divisible by K.
- We use a while loop to keep incrementing N until we find a number that is divisible by K.
- Once we find the smallest number greater than or equal to N that is divisible by K, we return it.
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 kint findNum(int N, int K){ while (N % K != 0) { N++; } return N;}// Driver codeint 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;} |
Output
Smallest number greater than or equal to 45 that is divisible by 6 is 48
Time Complexity: O(K), where K is the divisor. The while loop will run at most K times before finding a number that is divisible by K.
Space Complexity: O(1), because we are only using a constant amount of extra space to store the input variables
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 kint findNum(int N, int K){ int rem = (N + K) % K; if (rem == 0) return N; else return N + K - rem;}// Driver codeint 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 approachpublic 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 CodeN = 45K = 6print('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 approachpublic 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 kfunction 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)
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