Given a positive integers N, the task is to find the smallest N digit number divisible by N.
Examples:
Input: N = 2
Output: 10
Explanation:
10 is the smallest 2-digit number which is divisible by 2.Input: N = 3
Output: 102
Explanation:
102 is the smallest 3-digit number which is divisible by 3.
Naive Approach: The naive approach is to iterate from smallest N-digit number(say S) to largest N-digit number(say L). The first number between [S, L] divisible by N is the required result.
Below is the implementation of above approach:
C++
// C++ program for the above approach#include <iostream>#include <math.h>using namespace std;// Function to find the smallest// N-digit number divisible by Nvoid smallestNumber(int N){ // Find largest n digit number int L = pow(10, N) - 1; // Find smallest n digit number int S = pow(10, N - 1); for (int i = S; i <= L; i++) { // If i is divisible by N, // then print i and return ; if (i % N == 0) { cout << i; return; } }}// Driver Codeint main(){ // Given Number int N = 2; // Function Call smallestNumber(N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the smallest// N-digit number divisible by Nstatic void smallestNumber(int N){ // Find largest n digit number int L = (int) (Math.pow(10, N) - 1); // Find smallest n digit number int S = (int) Math.pow(10, N - 1); for (int i = S; i <= L; i++) { // If i is divisible by N, // then print i and return ; if (i % N == 0) { System.out.print(i); return; } }}// Driver Codepublic static void main(String[] args){ // Given Number int N = 2; // Function Call smallestNumber(N);}}// This code is contributed by Amit Katiyar |
Python3
# Python3 program for the above approach# Function to find the smallest# N-digit number divisible by Ndef smallestNumber(N): # Find largest n digit number L = pow(10, N) - 1; # Find smallest n digit number S = pow(10, N - 1); for i in range(S, L): # If i is divisible by N, # then print i and return ; if (i % N == 0): print(i); return; # Driver Code if __name__ == "__main__" : # Given number N = 2; # Function call smallestNumber(N)# This code is contributed by rock_cool |
C#
// C# program for the above approachusing System;class GFG{// Function to find the smallest// N-digit number divisible by Nstatic void smallestNumber(int N){ // Find largest n digit number int L = (int)(Math.Pow(10, N) - 1); // Find smallest n digit number int S = (int)Math.Pow(10, N - 1); for(int i = S; i <= L; i++) { // If i is divisible by N, // then print i and return ; if (i % N == 0) { Console.Write(i); return; } }}// Driver Codepublic static void Main(){ // Given number int N = 2; // Function call smallestNumber(N);}}// This code is contributed by Nidhi_biet |
Javascript
<script>// Javascript program for the above approach// Function to find the smallest// N-digit number divisible by Nfunction smallestNumber(N){ // Find largest n digit number let L = Math.pow(10, N) - 1; // Find smallest n digit number let S = Math.pow(10, N - 1); for(let i = S; i <= L; i++) { // If i is divisible by N, // then print i and return ; if (i % N == 0) { document.write(i); return; } }}// Driver code// Given Numberlet N = 2;// Function CallsmallestNumber(N); // This code is contributed by divyeshrabadiya07</script> |
Output:
10
Time Complexity: O(L – S), where L and S is the largest and smallest N-digit number respectively.
Auxiliary Space: O(1)
Efficient Approach: If the number divisible by N, then the number will be of the form N * X for some positive integer X.
Since it has to be smallest N-digit number, then X will be given by:
. Therefore, the smallest number N-digit number is given by:
For Example:
For N = 3, the smallest 3-digit number is given by:
=>=>
=>
=> 102
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <iostream>#include <math.h>using namespace std;// Function to find the smallest// N-digit number divisible by Nint smallestNumber(int N){ // Return the smallest N-digit // number calculated using above // formula return N * ceil(pow(10, (N - 1)) / N);}// Driver Codeint main(){ // Given N int N = 2; // Function Call cout << smallestNumber(N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the smallest// N-digit number divisible by Nstatic int smallestNumber(int N){ // Return the smallest N-digit // number calculated using above // formula return (int) (N * Math.ceil(Math.pow(10, (N - 1)) / N));}// Driver Codepublic static void main(String[] args){ // Given N int N = 2; // Function Call System.out.print(smallestNumber(N));}}// This code is contributed by Princi Singh |
Python3
# Python3 program for the above approachimport math# Function to find the smallest# N-digit number divisible by Ndef smallestNumber(N): # Return the smallest N-digit # number calculated using above # formula return N * math.ceil(pow(10, (N - 1)) // N);# Driver Code# Given NN = 2;# Function Callprint(smallestNumber(N));# This code is contributed by Code_Mech |
C#
// C# program for the above approachusing System;class GFG{// Function to find the smallest// N-digit number divisible by Nstatic int smallestNumber(int N){ // Return the smallest N-digit // number calculated using above // formula return (int) (N * Math.Ceiling(Math.Pow(10, (N - 1)) / N));}// Driver Codepublic static void Main(){ // Given N int N = 2; // Function Call Console.Write(smallestNumber(N));}}// This code is contributed by Code_Mech |
Javascript
<script> // Javascript program for the above approach // Function to find the smallest // N-digit number divisible by N function smallestNumber(N) { // Return the smallest N-digit // number calculated using above // formula return N * Math.ceil(Math.pow(10, (N - 1)) / N); } // Given N let N = 2; // Function Call document.write(smallestNumber(N)); // This code is contributed by divyesh072019.</script> |
Output:
10
Time Complexity: O(log(N))
Auxiliary Space: O(1)
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