Given 
and 
. The task is to find an N digit number that is divisible by D ( 2 <= D <= 10). If it is not possible, then print Impossible.
Examples:
Input : N = 2 and D = 2
Output : 20Input : N = 1 and D = 10
Output : Impossible
Approach: There are two conditions, D=10 and D not equal to 10. If D = 10 and N = 1 then the only answer is not possible and in all other conditions, the answer will be possible.
1. If D is 10, Print 1 followed by n-1 times zero. 2. If D is not 10 Print D followed by n-1 times zero
Below is the implementation of the above approach:
C++
// CPP program to Find N digits// number which is divisible by D#include <bits/stdc++.h>using namespace std;// Function to return N digits// number which is divisible by Dstring findNumber(int n, int d){ // to store answer string ans = ""; if (d != 10) { ans += to_string(d); for (int i = 1; i < n; i++) ans += '0'; } else { if (n == 1) ans += "Impossible"; else { ans += '1'; for (int i = 1; i < n; i++) ans += '0'; } } return ans;}// Driver codeint main(){ int n = 12, d = 3; cout << findNumber(n, d); return 0;} |
Java
// Java program to Find N digits// number which is divisible by Dimport java.io.*;class GFG {// Function to return N digits// number which is divisible by Dstatic String findNumber(int n, int d){ // to store answer String ans = ""; if (d != 10) { ans += Integer.toString(d); for (int i = 1; i < n; i++) ans += '0'; } else { if (n == 1) ans += "Impossible"; else { ans += '1'; for (int i = 1; i < n; i++) ans += '0'; } } return ans;}// Driver code public static void main (String[] args) { int n = 12, d = 3; System.out.println(findNumber(n, d)); }}// This code is contributed by anuj_67.. |
Python 3
# Python 3 program to Find N digits# number which is divisible by D# Function to return N digits# number which is divisible by Ddef findNumber(n, d): # to store answer ans = "" if (d != 10) : ans += str(d) for i in range(1,n): ans += '0' else : if (n == 1): ans += "Impossible" else : ans += '1' for i in range(1,n): ans += '0' return ans# Driver codeif __name__ == "__main__": n = 12 d = 3 print(findNumber(n, d)) # This code is contributed by# ChitraNayal |
C#
// C# program to Find N digits// number which is divisible by Dusing System;class GFG {// Function to return N digits// number which is divisible by Dstatic string findNumber(int n, int d){ // to store answer string ans = ""; if (d != 10) { ans += d.ToString(); for (int i = 1; i < n; i++) ans += '0'; } else { if (n == 1) ans += "Impossible"; else { ans += '1'; for (int i = 1; i < n; i++) ans += '0'; } } return ans;}// Driver codepublic static void Main () { int n = 12, d = 3; Console.WriteLine(findNumber(n, d));}}// This code is contributed by Subhadeep |
PHP
<?php// PHP program to Find N digits// number which is divisible by D// Function to return N digits// number which is divisible by Dfunction findNumber($n, $d){ // to store answer $ans = ""; if ($d != 10) { $ans .= strval($d); for($i = 1; $i < $n; $i++) $ans .= '0'; } else { if (n == 1) $ans .= "Impossible"; else $ans .= '1'; for($i = 1; $i < $n; $i++) $ans .= '0'; } return $ans;}// Driver code$n = 12;$d = 3;print(findNumber($n, $d)); // This code is contributed by mits |
Javascript
<script>// Javascript program to Find N digits// number which is divisible by D // Function to return N digits// number which is divisible by D function findNumber(n,d) { // to store answer let ans = ""; if (d != 10) { ans += (d).toString(); for (let i = 1; i < n; i++) ans += '0'; } else { if (n == 1) ans += "Impossible"; else { ans += '1'; for (let i = 1; i < n; i++) ans += '0'; } } return ans; } // Driver code let n = 12, d = 3; document.write(findNumber(n, d));// This code is contributed by avanitrachhadiya2155</script> |
Output:
300000000000
Time Complexity: O(n)
Auxiliary Space: O(n)
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