# Print a number containing K digits with digital root D

Given a digital root ‘D’ and number of digits ‘K’. The task is to print a number containing K digits that has its digital root equal to D. Print ‘-1’ if such a number does not exist.

**Examples:**

Input: D = 4, K = 4 Output: 4000 No. of digits is 4. Sum of digits is also 4. Input: D = 0, K = 1 Output: 0

**Approach:** A key observation to solving this problem is that appending any number of 0s to a number does not change its digital root. Hence **D followed by (K-1) 0’s** is a simple solution.

Special case when D is 0 and K is not 1 does not have a solution since the only number with digital root 0 is 0 itself.

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 a number ` `void` `printNumberWithDR(` `int` `k, ` `int` `d) ` `{ ` ` ` ` ` `// If d is 0 k has to be 1 ` ` ` `if` `(d == 0 && k != 1) ` ` ` `cout << ` `"-1"` `; ` ` ` ` ` `else` `{ ` ` ` `cout << d; ` ` ` `k--; ` ` ` ` ` `// Print k-1 zeroes ` ` ` `while` `(k--) ` ` ` `cout << ` `"0"` `; ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `k = 4, d = 4; ` ` ` ` ` `printNumberWithDR(k, d); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the above approach ` ` ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ` `// Function to find a number ` `static` `void` `printNumberWithDR(` `int` `k, ` `int` `d) ` `{ ` ` ` ` ` `// If d is 0 k has to be 1 ` ` ` `if` `(d == ` `0` `&& k != ` `1` `) ` ` ` `System.out.print( ` `"-1"` `); ` ` ` ` ` `else` `{ ` ` ` `System.out.print(d); ` ` ` `k--; ` ` ` ` ` `// Print k-1 zeroes ` ` ` `while` `(k-->` `0` `) ` ` ` `System.out.print( ` `"0"` `); ` ` ` `} ` `} ` ` ` `// Driver code ` ` ` ` ` `public` `static` `void` `main (String[] args) { ` ` ` `int` `k = ` `4` `, d = ` `4` `; ` ` ` ` ` `printNumberWithDR(k, d); ` ` ` `} ` `} ` ` ` ` ` `//This code is contributed by inder_verma.. ` |

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## Python3

`# Python3 implementation of ` `# the above approach ` ` ` `# Function to find a number ` `def` `printNumberWithDR( k, d ) : ` ` ` ` ` `# If d is 0, k has to be 1 ` ` ` `if` `d ` `=` `=` `0` `and` `k !` `=` `1` `: ` ` ` `print` `(` `-` `1` `, end ` `=` `"") ` ` ` ` ` `else` `: ` ` ` `print` `(d, end ` `=` `"") ` ` ` `k ` `-` `=` `1` ` ` ` ` `# Print k-1 zeroes ` ` ` `while` `k : ` ` ` ` ` `print` `(` `0` `,end ` `=` `"") ` ` ` `k ` `-` `=` `1` ` ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `k, d ` `=` `4` `, ` `4` ` ` ` ` `# Function call ` ` ` `printNumberWithDR( k, d ) ` ` ` `# This code is contributed by ` `# ANKITRAI1 ` |

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## C#

`// C# implementation of the above approach ` `using` `System; ` ` ` `class` `GFG { ` ` ` `// Function to find a number ` `static` `void` `printNumberWithDR(` `int` `k, ` `int` `d) ` `{ ` ` ` ` ` `// If d is 0 k has to be 1 ` ` ` `if` `(d == 0 && k != 1) ` ` ` `Console.Write( ` `"-1"` `); ` ` ` ` ` `else` `{ ` ` ` ` ` `Console.Write(d); ` ` ` `k--; ` ` ` ` ` `// Print k-1 zeroes ` ` ` `while` `(k-->0) ` ` ` `Console.Write( ` `"0"` `); ` ` ` `} ` `} ` ` ` `// Driver code ` `static` `public` `void` `Main () ` `{ ` ` ` `int` `k = 4, d = 4; ` ` ` ` ` `printNumberWithDR(k, d); ` `} ` `} ` ` ` `// This code is contributed by ajit. ` |

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## PHP

`<?php ` `// PHP implementation of the above approach ` ` ` `// Function to find a number ` `function` `printNumberWithDR(` `$k` `, ` `$d` `) ` `{ ` ` ` ` ` `// If d is 0 k has to be 1 ` ` ` `if` `(` `$d` `== 0 && ` `$k` `!= 1) ` ` ` `echo` `"-1"` `; ` ` ` ` ` `else` ` ` `{ ` ` ` `echo` `$d` `; ` ` ` `$k` `--; ` ` ` ` ` `// Print k-1 zeroes ` ` ` `while` `(` `$k` `--) ` ` ` `echo` `"0"` `; ` ` ` `} ` `} ` ` ` `// Driver code ` `$k` `= 4; ` `$d` `= 4; ` ` ` `printNumberWithDR(` `$k` `, ` `$d` `); ` ` ` `// This code is contributed ` `// by akt_mit ` `?> ` |

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**Output:**

4000

** Time complexity:** O(K)

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